Tarun Kumar Rawat
Signals and Systems
Chapter-1 Signals and Systems 1.1 Prove each of the following statements. In each case, let energy signal x1 (t) have energy E[x1 (t)], let energy signal x2 (t) have energy E[x2 (t)], and let T be a nonzero, finite, real-valued constant. (i) Prove E[T x1 (t)] = T 2 E[x1 (t)]. That is, amplitude scaling a signal by constant T scales the signal energy by T 2 . (ii) Prove E[x1 (t)] = E[x1 (t − T )]. That is, shifting a signal does not affect its energy. (iii) If (x1 (t) 6= 0) ⇒ (x2 (t) = 0) and (x2 (t) 6= 0) ⇒ (x1 (t) = 0), then prove E[x1 (t) + x2 (t)] = E[x1 (t)] + E[x2 (t)]. That is, the energy of the sum of two nonoverlapping signals is the sum of the two individual energies. (iv) Prove E[x1 (T t)] = 1/|T |E[x1 (t)]. That is, time-scaling a signal by T reciprocally scales the signal energy by 1/|T |. Solution 1.1 Given that Z E[x1 (t)] = Ex1 =
∞
|x1 (t)|2 dt
−∞
(i) The energy contained in the signal T x1 (t) is given by Z ∞ Z E[T x1 (t)] = |T x1 (t)|2 dt = T 2 −∞
∞ −∞
|x1 (t)|2 dt = T 2 Ex1
(ii) The energy contained in the signal, x1 (t − T ), is given by Z ∞ E[x1 (t − T )] = |x1 (t − T )|2 dt −∞
A change of variables is performed by letting t − T = τ , which also yields dt = dτ , τ = −∞ as t = −∞, and τ = ∞ as t = ∞. Therefore, Z ∞ E[x1 (t − T )] = |x(τ )|2 dτ = E[x1 (t)] = Ex1 −∞
(iii) The energy contained in the signal x1 (t) + x2 (t) is given by Z ∞ Z ∞ Z 2 2 E[x1 (t) + x2 (t)] = |x1 (t) + x2 (t)| dt = [x1 (t) + x2 (t)] dt = −∞
−∞
−∞
Since x1 (t) and x2 (t) are nonoverlapping signals, i.e. x1 (t)x2 (t) = 0, we have Z ∞ Z ∞ Z 2 2 2 E[x1 (t) + x2 (t)] = [x1 (t) + x2 (t)]dt = [x1 (t)]dt + −∞
∞
−∞
∞
−∞
[x21 (t) + x22 (t) + 2x1 (t)x2 (t)]dt
[x22 (t)]dt = Ex1 + Ex2
(iv) Case-I: T > 0. The energy contained in the signal, x1 (T t), is given by Z ∞ Z ∞ 2 E[x1 (T t)] = |x1 (T t)| dt = x21 (T t)dt −∞
−∞
A change of variables is performed by letting T t = τ , which also yields dt = T1 dτ , τ = −∞ as t = −∞, and τ = ∞ as t = ∞. Therefore, Z Ex1 1 ∞ |x1 (τ )|2 dτ = E[x1 (T t)] = T −∞ T Case-II: T < 0. The energy contained in the signal, x1 (−T t), is given by Z ∞ Z ∞ 2 E[x1 (−T t)] = |x1 (−T t)| dt = x21 (−T t)dt −∞
−∞
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Signals and Systems
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A change of variables is performed by letting −T t = τ , which also yields dt = − T1 dτ , τ = ∞ as t = −∞, and τ = −∞ as t = ∞. Therefore, Z Z 1 ∞ Ex1 1 −∞ |x1 (τ )|2 dτ = |x1 (τ )|2 dτ = E[x1 (T t)] = − T ∞ T −∞ T Hence, E[x1 (T t)] = 1/|T |E[x1 (t)]. 1.2 Show that the power of a signal x(t) =
n X
Ck ejωk t
is Px =
k=m
n X
|Ck |2
k=m
assuming all frequencies to be distinct, that is, ωi 6= ωk for all i 6= k.
Solution 1.2 The power contained in the signal x(t) is given by 1 Px = lim T →∞ T
Z
T /2
1 |x(t)| dt = lim T →∞ T −T /2
Z
2
= lim
T →∞
1 T
1 T →∞ T
= lim
Z Z
¯ n ¯2 ¯X ¯ ¯ jωk t ¯ Ck e ¯ ¯ dt ¯ −T /2 ¯ T /2
k=m
T /2
n X
−T /2 k=m n T /2 X −T /2 k=m
¯ ¯2 |Ck |2 ¯ejωk t ¯ dt n n X 1 X T |Ck |2 = |Ck |2 T →∞ T
|Ck |2 dt = lim
1.3 For the signal x(t) illustrated in Fig.1.35(a), sketch (i) x(t − 4) (ii) x(t/1.5) (iii) x(−t) (iv) x(2t − 4) (v) x(2 − t)
Solution 1.3
Fig.S1.1
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Signals and Systems
1.4
3
In Fig.1.42, express signals x1 (t), x2 (t) and x3 (t) in terms of signal x(t) and its time-shifted, time-scaled, or time-reversed versions.
Solution 1.4
x1 (t) = x(t + 1) + x(−t + 1) The signal x(t + 1) and x(−t + 1) are shown in Fig. S1.2(a). Now, define a signal y(t) = x(t) + x(−t + 1). x2 (t) = y(t) + y(−t) = x(t) + x(−t + 1) + x(−t) + x(t + 1) The signal y(t) and y(−t) ¡ ¢ in Fig S1.2(b). ¡ ¢are shown Define a signal z(t) = y 2t + x 2t . x3 (t) = z(t) + z(−t) µ ¶ µ ¶ µ ¶ µ ¶ t t t t +x +y − +x − =y 2 2 2 2 µ µ ¶ µ ¶ µ ¶ ¶ µ ¶ ¶ µ t t t t t t +x − +1 +x +x − +x +1 +x − x3 (t) = x 2 2 2 2 2 2 ¶ µ µ ¶ µ ¶ µ ¶ t t t t + 2x − +x − +1 +x +1 x3 (t) = 2x 2 2 2 2 The signal z(t) and z(−t) are shown in Fig S1.2(c).
Fig. S1.2
1.5³ Simplify the following expressions: ´ (i) (ii)
sin t 2 +2
³t
ω+2 ω 2 +9 −t
δ(t) ´ δ(ω)
o (iii) ³ [e cos(3t ´ − 60 )]δ(t) 1 δ(ω + 3) (iv) jω+2
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Tarun Kumar Rawat
(v)
¡ sin kω ¢ ω
Signals and Systems
δ(ω)
Solution 1.5 Since x(t)δ(t) = x(0)δ(t), we have (i) µ
sin t t2 + 2
¶
µ δ(t) =
sin 0 0+2
¶ δ(t) = 0
(ii) µ
ω+2 ω2 + 9
¶
µ δ(ω) =
0+2 0+9
¶ δ(ω) =
2 δ(ω) 9
(iii) [e−t cos(3t − 60o )]δ(t) = [e0 cos(0 − 60o )]δ(t) = cos(−60o )δ(t) =
1 δ(t) 2
(iv) µ
1 jω + 2
¶
µ δ(ω + 3) =
1 −3j + 2
¶ δ(ω + 3)
(v) µ
sin kω ω
¶
µ δ(ω) =
sin 0 0
¶ δ(ω) = δ(ω)
1.6R ∞Evaluate the ¢ following integrals: ¡
(i) −∞ 32 t − 23 δ(t − 1)dt ¢ ¡ R∞ (ii) −∞ (t − 1)δ 32 t − 23 dt ¢ ¡ ¢¤ ¡ R −2 £ 3 (iii) −3 e−t+1 + sin 2π 3 t δ t − 2 dt ¢ ¡ ¢¤ ¡ R2 £ t δ t − 23 dt (iv) −3 e−t+1 + sin 2π 3 R∞ (v) −∞ e−5t+1 δ 0 (t − 5)dt
Solution 1.6 (i) Z
∞
µ
3 2 t− 3 2
−∞
¡ ¢ (ii) Since δ(at + b) = δ a(t + b/a) = Z
µ
∞
(t − 1)δ −∞
(iii) Since t =
3 2
3 2 t− 3 2
1 |a| δ(t
¶
¶
· δ(t − 1)dt =
2 3 t− 3 2
¸ = t=1
2 3 5 − =− 3 2 6
+ b/a), we have
µ ¶ µ ¶ ¯ 3 9 3 9 3 15 ¯ dt = (t − 1) δ t − dt = (t − 1)¯ 9 = −1 = 2 4 2 2 4 8 t= −∞ 4 Z
∞
is not in the interval −3 < t < −2, the shifting property of the impulse function yields ¶¸ µ ¶ µ Z −2 · 3 2π −t+1 t δ t− dt = 0 e + sin 3 2 −3
(iv) Z
2
·
µ e
−t+1
−3
+ sin
2π t 3
¶¸ µ ¶ · ¶¸ µ 3 2π −t+1 δ t− t dt = e + sin 2 3 t= 3 ¶¸ 2 · µ 3 1 1 2π 3 = [e− 2 + sin(π)] = e− 2 = e− 2 +1 + sin 3 2
(v) Z
Z
∞
e −∞
−5t+1 0
∞
δ (t − 5)dt = − −∞
d −5t+1 e δ(t − 5)dt = 5 dt
Z
∞ −∞
¯ e−5t+1 δ(t − 5)dt = 5e−5t+1 ¯t=5 = 5e−24
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1.7 (a) If xe (t) and x0 (t) are even and odd components of a real signal x(t), then show that Z
∞
xe (t)x0 (t)dt = 0 −∞
(b) Show that Z
Z
∞
∞
x(t)dt = −∞
xe (t)dt −∞
Solution 1.7 (a) Consider the LHS Z
Z
∞
∞
µ
xe (t)x0 (t)dt = −∞
−∞ Z ∞
x(t) + x(−t) 2
¶µ
x(t) − x(−t) 2
1 [x2 (t) − x2 (−t)]dt 4 −∞ ·Z ∞ ¸ Z ∞ 1 x2 (t)dt − x2 (−t)dt = 4 −∞ −∞ ·Z ∞ ¸ Z ∞ 1 x2 (t)dt − x2 (τ )dτ = 4 −∞ −∞ ·Z ∞ ¸ Z ∞ 1 x2 (t)dt − x2 (t)dt = 0 = 4 −∞ −∞
=
(b) Consider the LHS Z
Z
∞
∞
x(t)dt = −∞
−∞ Z ∞
=
[xe (t) + xo (t)]dt Z ∞ xe (t)dt + xo (t)dt
−∞
−∞
Using property no.3 (page. 52) of even and odd signals, we have Z ∞ Z ∞ Z x(t)dt = xe (t)dt + 0 = −∞
−∞
∞
xe (t)dt
−∞
1.8 Find and sketch the even and the odd components of the following: (i) u(t)
(ii) r(t) = tu(t)
(iii) sin(ω0 t)u(t)
(iv) cos(ω0 t)u(t)
Solution 1.8 (i) See part (a) of Example 1.24. (ii)
( t t≥0 r(t) = tu(t) = 0 t<0 Replacing t with −t in the above expression, we get ( r(−t) = −tu(−t) =
−t 0
−t ≥ 0 −→ t ≤ 0 −t < 0 −→ t > 0
Applying Eqs. (1.59) and (1.51) to get the even and odd part of r(t) respectively. ( t t≥0 1 re (t) = [r(t) + r(−t)] = 2−t 2 t<0 2 and 1 ro (t) = [r(t) − r(−t)] = 2
(
t 2 t 2
t≥0 t<0
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The signal r(t), r(−t), re (t), and r0 (t) are shown in Fig. S1.3(a). (iii) Consider the given signal ( sin(ω0 t) x(t) = sin(ω0 t)u(t) = 0
t>0 t<0
Replacing t with −t in the above expression, we get ( x(−t) = sin(−ω0 t)u(−t) =
sin(−ω0 t) 0
−t > 0 −→ t < 0 −t < 0 −→ t > 0
Applying Eqs. (1.59) and (1.51) to get the even and odd part of x(t) respectively. ( 1 sin(ω0 t) 1 1 1 xe (t) = [x(t) + x(−t)] = sin(ω0 t)u(t) + sin(−ω0 t)u(−t) = 21 2 2 2 2 sin(−ω0 t)
t>0 t<0
and xo (t) =
1 1 1 [x(t) − x(−t)] = sin(ω0 t)u(t) − sin(−ω0 t)u(−t) 2 2 2 1 1 = sin(ω0 t)u(t) + sin(ω0 t)u(−t) 2 2 1 1 = sin(ω0 t)[u(t) + u(−t)] = sin(ω0 t) 2 2
The signals sin(ω0 t)u(t), sin(−ω0 t)u(−t), xe (t) and x0 (t) are shown in Fig. S1.3(b). (iv) Consider the given signal ( cos(ω0 t) t > 0 x(t) = cos(ω0 t)u(t) = 0 t<0 Replacing t with −t in the above expression, we get ( cos(ω0 t) −t > 0 −→ t < 0 x(−t) = cos(−ω0 t)u(−t) = cos(ω0 t)u(−t) = 0 −t < 0 −→ t > 0 Applying Eqs. (1.59) and (1.51) to get the even and odd part of x(t) respectively. xe (t) =
1 1 1 1 1 [x(t) + x(−t)] = cos(ω0 t)u(t) + cos(ω0 t)u(−t) = cos(ω0 t)[u(t) + u(−t)] = cos(ω0 t) 2 2 2 2 2
and 1 1 1 xo (t) = [x(t) − x(−t)] = cos(ω0 t)u(t) − cos(ω0 t)u(−t) = 2 2 2
(
1 2 cos(ω0 t) − 21 cos(ω0 t)
The signals cos(ω0 t)u(t), cos(ω0 t)u(−t), xe (t) and x0 (t) are shown in Fig. S1.3(c).
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Fig. S1.3
1.9 An aperiodic signal is defined as x(t) = sin(πt)u(t), where u(t) is the continuous-time step function. Is the odd portion of this signal, x0 (t), periodic? Justify your answer. Solution 1.9 Consider the given signal ( sin(πt) t > 0 x(t) = sin(πt)u(t) = 0 t<0 Replacing t with −t in the above expression, we get
(
x(−t) = sin(−πt)u(−t) =
sin(−πt) −t > 0 −→ t < 0 0 −t < 0 −→ t > 0
The odd part of x(t) is given by xo (t) = = = = xo (t) =
1 [x(t) − x(−t)] 2 1 1 sin(πt)u(t) − sin(−πt)u(−t) 2 2 1 1 sin(πt)u(t) + sin(πt)u(−t) 2 2 1 sin(πt)[u(t) + u(−t)] 2 1 sin(πt) 2
2π The odd part xo (t) = 21 sin(πt) is periodic with period T = 2π ω = π = 2. 1.10 The output of the system, y(t), is related to the input x(t) as Z t ³ ´ y(t) = cos 2πfc t + k x(τ )dτ −∞
where k is a constant parameter. (a) Show that the system is nonlinear. (b) Show that the system is time-variant. Solution 1.10 (a) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (t) and x2 (t). Z t ³ ´ x1 (t) → y1 (t) = cos 2πfc t + k x1 (τ )dτ Z
³
−∞ t
x2 (t) → y2 (t) = cos 2πfc t + k Z
³
´ x2 (τ )dτ
−∞ t
x3 (t) → y3 (t) = cos 2πfc t + k
´ x3 (τ )dτ
−∞
Let x3 (t) be a linear combination of x1 (t) and x2 (t). That is, x3 (t) = ax1 (t) + bx2 (t) where a and b are arbitrary constants. If the system is linear, then y3 (t) = ay1 (t) + by2 (t) Consider the LHS of the above equation Z ³ y3 (t) = cos 2πfc t + k ³
Z
x3 (τ )dτ −∞ t
= cos 2πfc t + k ³
´
t
´ [ax1 (τ ) + bx2 (τ )]dτ
−∞ Z t
= cos 2πfc t + a k
Z
−∞
´
t
x1 (τ )dτ + b k
x2 (τ )dτ −∞
6= ay1 (t) + by2 (t)
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Since LHS 6= RHS, we conclude that the system is nonlinear. (b) Let x1 (t) be an arbitrary input to the system, and let Z ³ y1 (t) = cos 2πfc t + k
´
t
x1 (τ )dτ −∞
be the corresponding output. Then consider a second input obtained by shifting x1 (t) in time: x2 (t) = x1 (t − t0 ) The output corresponding to this input is Z ³ y2 (t) = cos 2πfc t + k Z
³
x2 (τ )dτ −∞ t
= cos 2πfc t + k
´ x1 (τ − t0 )dτ
Z
³
´
t
−∞ t−t0
y2 (t) = cos 2πfc t + k
´ x1 (τ )dτ
−∞
Now, consider Z ³ y1 (t) = cos 2πfc t + k ³
´
t
x1 (τ )dτ −∞
Z
´
t−t0
y1 (t − t0 ) = cos 2πfc (t − t0 ) + k
x1 (τ )dτ −∞
6= y2 (t) We see that y2 (t) 6= y1 (t − t0 ), and therefore, this system is time-variant. 1.11 The system that follow have input x(n) and output y(n). For each system, determine whether it is (i) memoryless, (ii) stable, (iii) causal, (iv) linear, and (v) time-invariant. (a) y(n) = loge [x(n)] (b) y(n) = log10 [|x(n)|] (c) y(n) = median{x(n − 1), x(n), x(n + 1)} (d) y(n) = Trun[x(n)], where Trun[x(n)] denotes the integer part of x(n) obtained by truncation. (e) y(n) = Round[x(n)], where Round[x(n)] denotes the integer part of x(n) obtained by rounding. (f) y(n) = |x(n)| (g) y(n) = sgn[x(n)] Solution 1.11 (a) (i) Consider the given input-output relation of the system. y(n) = loge [x(n)] y(0) = loge [x(0)] y(1) = loge [x(1)] y(−1) = loge [x(−1)] The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Assume that the input signal x(n) satisfies the condition |x(n)| ≤ Bx < ∞ for all n. We then find that |y(n)| = |loge [x(n)]| = |loge [Bx ]| = By < ∞ We conclude that in this system if any input is bounded by an arbitrary positive number Bx , the corresponding output is bounded. Thus, the given system is stable.
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(iii) Consider the given input-output relation of the system. y(n) = loge [x(n)] y(0) = loge [x(0)] y(1) = loge [x(1)] y(−1) = loge [x(−1)] The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) and x2 (n). x1 (n) → y1 (n) = loge [x1 (n)] x2 (n) → y2 (n) = loge [x2 (n)] x3 (n) → y3 (n) = loge [x3 (n)] Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = ax1 (n) + bx2 (n) where a and b are arbitrary scalars. If the system is linear, then y3 (n) = ay1 (n) + by2 (n) Consider the LHS of the above equation y3 (n) = loge [x3 (n)] = loge [ax1 (n) + bx2 (n)] 6= ay1 (n) + by2 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = loge [x1 (n)] be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = loge [x2 (n)] = loge [x1 (n − n0 )] Now, consider y1 (n) = loge [x1 (n)] y1 (n − n0 ) = loge [x1 (n − n0 )] Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. (b) (i) Consider the given input-output relation of the system. y(n) = log10 [|x(n)|] y(0) = log10 [|x(0)|] y(1) = log10 [|x(1)|] y(−1) = log10 [|x(−1)|] The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Assume that the input signal x(n) satisfies the condition |x(n)| ≤ Bx < ∞ for all n.
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We then find that |y(n)| = |log10 [|x(n)|]| = |log10 [|Bx |]| = By < ∞ We conclude that in this system if any input is bounded by an arbitrary positive number Bx , the corresponding output is bounded. Thus, the given system is stable. (iii) Consider the given input-output relation of the system. y(n) = log10 [|x(n)|] y(0) = log10 [|x(0)|] y(1) = log10 [|x(1)|] y(−1) = log10 [|x(−1)|] The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) and x2 (n). x1 (n) → y1 (n) = log10 [|x1 (n)|] x2 (n) → y2 (n) = log10 [|x2 (n)|] x3 (n) → y3 (n) = log10 [|x3 (n)|] Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = ax1 (n) + bx2 (n) where a and b are arbitrary constants. If the system is linear, then y3 (n) = ay1 (n) + by2 (n) Consider the LHS of the above equation y3 (n) = log10 [|x3 (n)|] = log10 [|ax1 (n) + bx2 (n)|] 6= ay1 (n) + by2 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = log10 [|x1 (n)|] be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = log10 [|x2 (n)|] = log10 [|x1 (n − n0 )|] Now, consider y1 (n) = log10 [|x1 (n)|] y1 (n − n0 ) = log10 [|x1 (n − n0 )|] Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. (c) Median filtering is implemented by sliding a window of odd length over the input sequence x(n) one sample at a time. At the nth instant, the input samples inside the window are rank ordered from the smallest to the largest in
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values, and the sample at the middle is the median value. For example, median{2, −3, 7, −10, 5} = 5. (i) Consider the given input-output relation of the system. y(n) = median{x(n − 1), x(n), x(n + 1)} To calculate the median value, the past, present and future inputs are required. Consequently, the present output depends on the present, past and future inputs. We may conclude, this system is a with-memory (dynamic) system. (ii) For a bounded input x(n), the output y(n) of this system is always bounded. Thus, the given system is stable. (iii) To calculate the median value, the past, present and future inputs are required. Consequently, the present output depends on the present, past and future inputs. We may conclude, this system is a noncausal system. (iv) Median filtering is a nonlinear operation. Consider the following sequences as the input to a median filter: x1 (n) = {3, 5, 9}
and
x2 (n) = {2, −2, −2}
The corresponding outputs of the median filter are y1 (n) = median{3, 5, 9} = 5
and
y2 (n) = median{2, −2, −2} = −2
Now consider another input sequence x3 (n) = x1 (n) + x2 (n) = {5, 3, 7} Then the corresponding output is y3 (n) = median{5, 3, 7} = 5 On the other hand y1 (n) + y2 (n) = 3 6= y3 (n) Hence median filtering is a nonlinear operation. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = median{x1 (n − 1), x1 (n), x1 (n + 1)} be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = median{x2 (n − 1), x2 (n), x2 (n + 1)} = median{x1 (n − n0 − 1), x1 (n − n0 ), x1 (n − n0 + 1)} Now, consider y1 (n) = median{x1 (n − 1), x1 (n), x1 (n + 1)} y1 (n − n0 ) = median{x1 (n − n0 − 1), x1 (n − n0 ), x1 (n − n0 + 1)} Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. (d) (i) Consider the given input-output relation of the system. y(n) = Trun[x(n)] y(0) = Trun[x(0)] y(1) = Trun[x(1)] y(−1) = Trun[x(−1)] The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Assume that the input signal x(n) satisfies the condition |x(n)| ≤ Bx = 5.3 < ∞
for all n.
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Signals and Systems
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We then find that |y(n)| = |Trun[x(n)]| = |Trun[5.3]| = 5 = By < ∞ We conclude that in this system if any input is bounded by an arbitrary positive number Bx , the corresponding output is bounded. Thus, the given system is stable. (iii) Consider the given input-output relation of the system. y(n) = Trun[x(n)] y(0) = Trun[x(0)] y(1) = Trun[x(1)] y(−1) = Trun[x(−1)] The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) = 5.3 and x2 (n) = 6.8. x1 (n) → y1 (n) = Trun[x1 (n)] = Trun[5.3] = 5 x2 (n) → y2 (n) = Trun[x2 (n)] = Trun[6.8] = 6 x3 (n) → y3 (n) = Trun[x3 (n)] Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = x1 (n) + x2 (n) = 5.3 + 6.8 = 12.1 If the system is linear, then y3 (n) = y1 (n) + y2 (n) Consider the LHS of the above equation y3 (n) = Trun[x3 (n)] = Trun[x1 (n) + x2 (n)] = Trun[12.1] = 12 On the other hand y1 (n) + y2 (n) = 5 + 6 = 11 6= y3 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = Trun[x1 (n)] be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = Trun[x2 (n)] = Trun[x1 (n − n0 )] Now, consider y1 (n) = Trun[x1 (n)] y1 (n − n0 ) = Trun[x1 (n − n0 )] Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant.
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Tarun Kumar Rawat
Signals and Systems
13
(e) (i) Consider the given input-output relation of the system. y(n) = Round[x(n)] y(0) = Round[x(0)] y(1) = Round[x(1)] y(−1) = Round[x(−1)] The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Assume that the input signal x(n) satisfies the condition |x(n)| ≤ Bx = 5.7 < ∞
for all n.
We then find that |y(n)| = |Round[x(n)]| = |Round[5.7]| = 6 = By < ∞ We conclude that in this system if any input is bounded by an arbitrary positive number Bx , the corresponding output is bounded. Thus, the given system is stable. (iii) Consider the given input-output relation of the system. y(n) = Round[x(n)] y(0) = Round[x(0)] y(1) = Round[x(1)] y(−1) = Round[x(−1)] The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) = 5.3 and x2 (n) = 6.4. x1 (n) → y1 (n) = Round[x1 (n)] = Round[5.3] = 5 x2 (n) → y2 (n) = Round[x2 (n)] = Round[6.4] = 6 x3 (n) → y3 (n) = Round[x3 (n)] Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = x1 (n) + x2 (n) = 5.3 + 6.4 = 11.7 If the system is linear, then y3 (n) = y1 (n) + y2 (n) Consider the LHS of the above equation y3 (n) = Round[x3 (n)] = Round[x1 (n) + x2 (n)] = Round[11.7] = 12 On the other hand y1 (n) + y2 (n) = 5 + 6 = 11 6= y3 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = Round[x1 (n)] be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 )
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Tarun Kumar Rawat
Signals and Systems
14
The output corresponding to this input is y2 (n) = Round[x2 (n)] = Round[x1 (n − n0 )] Now, consider y1 (n) = Round[x1 (n)] y1 (n − n0 ) = Round[x1 (n − n0 )] Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. (f ) (i) Consider the given input-output relation of the system. y(n) = |x(n)| y(0) = |x(0)| y(1) = |x(1)| y(−1) = |x(−1)| The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Assume that the input signal x(n) satisfies the condition |x(n)| ≤ Bx < ∞
for all n.
We then find that y(n) = |[x(n)]| = By = Bx < ∞ We conclude that in this system if any input is bounded by an arbitrary positive number Bx , the corresponding output is bounded. Thus, the given system is stable. (iii) Consider the given input-output relation of the system. y(n) = |x(n)| y(0) = |x(0)| y(1) = |x(1)| y(−1) = |x(−1)| The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) = 5 and x2 (n) = −6. x1 (n) → y1 (n) = |x1 (n)| = |5| = 5 x2 (n) → y2 (n) = |x2 (n)| = | − 6| = 6 x3 (n) → y3 (n) = |x3 (n)| Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = x1 (n) + x2 (n) = 5 − 6 = −1 If the system is linear, then y3 (n) = y1 (n) + y2 (n) Consider the LHS of the above equation y3 (n) = |x3 (n)| = |x1 (n) + x2 (n)| = | − 1| = 1
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Tarun Kumar Rawat
Signals and Systems
15
On the other hand y1 (n) + y2 (n) = 5 + 6 = 11 6= y3 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = |x1 (n)| be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = |x2 (n)| = |x1 (n − n0 )| Now, consider y1 (n) = |x1 (n)| y1 (n − n0 ) = |x1 (n − n0 )| Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. (g) ( y(n) = sgn[x(n)] =
1 −1
x(n) > 0 x(n) < 0
(i) Consider the given input-output relation of the system. y(n) = sgn[x(n)] y(0) = sgn[x(0)] y(1) = sgn[x(1)] y(−1) = sgn[x(−1)] The present output depends on the present input only. We may conclude, this system is a memoryless (static) system. (ii) Since ( 1 x(n) > 0 y(n) = sgn[x(n)] = −1 x(n) < 0 We conclude that in this system if any input is bounded, the corresponding output is always bounded. Thus, the given system is stable. (iii) Consider the given input-output relation of the system. y(n) = sgn[x(n)] y(0) = sgn[x(0)] y(1) = sgn[x(1)] y(−1) = sgn[x(−1)] The present output depends on the present input only. We may conclude, this system is a causal system. (iv) To determine whether or not a system is linear, we consider two arbitrary inputs x1 (n) = 5 and x2 (n) = −3. x1 (n) → y1 (n) = sgn[x1 (n)] = sgn[5] = 1 x2 (n) → y2 (n) = sgn[x2 (n)] = sgn[−3] = −1 x3 (n) → y3 (n) = sgn[x3 (n)]
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Tarun Kumar Rawat
Signals and Systems
Let x3 (n) be a linear combination of x1 (n) and x2 (n). That is, x3 (n) = x1 (n) + x2 (n) = 5 − 3 = 2 If the system is linear, then y3 (n) = y1 (n) + y2 (n) Consider the LHS of the above equation y3 (n) = sgn[x3 (n)] = sgn[x1 (n) + x2 (n)] = sgn[2] = 1 On the other hand y1 (n) + y2 (n) = 1 − 1 = 0 6= y3 (n) Since LHS 6= RHS, we conclude that the system is non-linear. (v) Let x1 (n) be an arbitrary input to the system, and let y1 (n) = sgn[x1 (n)] be the corresponding output. Then consider a second input obtained by shifting x1 (n) in time: x2 (n) = x1 (n − n0 ) The output corresponding to this input is y2 (n) = sgn[x2 (n)] = sgn[x1 (n − n0 )] Now, consider y1 (n) = sgn[x1 (n)] y1 (n − n0 ) = sgn[x1 (n − n0 )] Since y2 (n) = y1 (n − n0 ), thus this system is time-invariant. 1.12 Sketch the following signals: (i) x1 (t) = u(t) + 5u(t − 1) − 2u(t − 2) (ii) x2 (t) = r(t) − r(t − 1) − u(t − 2) (iii) x3 (t) = 2u(t) + δ(t − 1) (iv) x4 (t) = u(t)u(t − 2) (v) x1 (t)x2 (t + 21 )
Solution 1.12
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Tarun Kumar Rawat
Signals and Systems
17
Fig. S1.4
1.13 A discrete-time signal is shown in Fig.1.43.
Sketch and label each of the following signals:
(i) x(n − 4) (ii) x(3 − n) (iii) x(3n) (iv) x(3n + 1) (v) x(n)u(3 − n) (vi) x(n − 2)δ(n − 2) ( vii) 12 x(n) + 21 (−1)n x(n) (viii) x(n − 1)2
Solution 1.13
Fig. S1.5
1.14 Find the derivative of the waveforms in Fig.1.40 and write the equations for the derivatives using shifted step and/or impulse functions. Solution 1.14 (a) 2E E E 2E r(t) − r(t − T ) + r(t − 3T ) − r(t − 4T ) T T T T 2E E dx1 (t) E 2E = x01 (t) = u(t) − u(t − T ) + u(t − 3T ) − u(t − 4T ) dt T T T T x1 (t) =
x01 (t) is shown in Fig. S1.6(a) (b) x2 (t) = r(t) − r(t − 2) − u(t − 2) dx2 (t) = x02 (t) = u(t) − u(t − 2) − δ(t − 2) dt x02 (t) is shown in Fig. S1.6(b) (c) x3 (t) = 1.5u(t) + 0.5u(t − 1) − 2u(t − 2) dx3 (t) 0 x3 (t) = 1.5δ(t) + 0.5δ(t − 1) − 2δ(t − 2) = x03 (t) is shown in Fig. S1.6(c)
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Tarun Kumar Rawat
Signals and Systems
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(d) 1 x4 (t) = u(t) + r(t) − 2 dx4 (t) 1 0 = x4 (t) = δ(t) + u(t) − dt 2
1 r(t − 2) − 2u(t − 2) 2 1 u(t − 2) − 2δ(t − 2) 2
x04 (t) is shown in Fig. S1.6(d)
Fig. S1.6
1.15 Show that if x(n) is periodic with period N , then n X
n+N X
x(k) =
k=n0
x(k)
k=n0 +N
Solution 1.15 Consider the RHS n+N X
x(k)
k=n0 +N
A change of variables is performed by letting k − N = m, which also yields k = m + N , m = n0 as k = n0 + N , and m = n as k = n + N . Therefore, n X
x(m + N )
m=n0
Since x(m + N ) = x(m), we have n X m=n0
x(m) =
n X
x(k) = LHS
k=n0
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