Signals Ty s

Electrical EngineeringSignals and Systems

WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK

2016

Detailed Explanations ofTry Yourself Questions

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T1 : Solution

(a)(a)(a)(a)(a)

As we know even part of any signal is � � � �

�

� �+ −x x

So if we use the above formulae we can determine that even part of x(t) is represented as signal given inoption (a).

T2 : Solution

(b)(b)(b)(b)(b)The signal x(t) = u(–t + 5) – u(–t – 4)

x( )t

t+5–4

1

Now we need to find y(t) = x(–2t + 2)To find y(t) first of all we will find x(t + 2) i.e. left shift signal x(t) by 2, then we will find x(2t + 2) i.e. scalingthe signal x(t + 2) by factor of 2 and then we will find x(–2t + 2) i.e. time reversal of signal x(2t + 2)

x( )t + 2

3–6

x( )t2 + 2

1.5–3

y t( ) = 2 + 2x(– )t

3–1.5

⇒⇒

Introduction1

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3Try yourself

T3 : Solution

(a)(a)(a)(a)(a)

The expression of x(t) is � � � � � � � ��

� � � � �∞

= −∞= δ − −δ − −∑x

So x(t) is a subtraction of two signals each periodic with period 4. So x(t) is periodic with period 4.

T6 : Solution

The signal is, x(t) = 3e–t u(t)

Now, energy of signal will be Ex =∞

− =∫ �

�

� ��

T7 : Solution

The signal x(t) is (t – 1)2

We need to find � � � ��

∞

−∞

δ − =∫ x x

So, x(1) = (1 – 1)2 = 0So answer is ‘0’.


T1 : Solution

(a)(a)(a)(a)(a)Given that Fourier series coefficient of x(t) is ak

So, x(t) ��⎯⎯⎯→ ak

Now, real part of x(t) is � ��

�

� �∗x x

and if x(t) ��⎯⎯⎯→ ak

then x∗(t) ��⎯⎯⎯→ ��∗−

So real part of x(t),� ��

�

� �∗x x ��⎯⎯⎯→

�

� �� ∗−+

T2 : Solution

(c)(c)(c)(c)(c)We know that Fourier series cannot be defined if the signal is non periodic. From the given options, signalgiven in option (c) is non periodic. So answer is (c).

T3 : Solution

(d)(d)(d)(d)(d)

For real periodic signal the Fourier series coefficients (ck) are conjugate symmetric � �� ∗−= which means

even magnitude and odd phase. So option (c).

Fourier Series2


5Try yourself

T4 : Solution

(d)(d)(d)(d)(d)

The signal x(t) is such that x(t) = ��

�� +⎛ ⎞− ⎜ ⎟⎝ ⎠x so the signal has half wave symmetry. So the signal will have

only odd harmonics, and since signal is even it will have only cosine terms.

T5 : Solution

Power of signals is ��

∞

−∞∑ ⇒

��

��

−∑

So power is =�

� � � � �

�

�� −

= + + + =∑


T1 : Solution

(b)(b)(b)(b)(b)The Fourier transform is X(ω) = u(ω) – u(ω – 2), we know that• If signal is real then X(ω) is conjugate symmetric.• If signal is imaginary then X(ω) is conjugate anti-symmetricThe given X(ω) is neither conjugate symmetric nor conjugate anti-symmetric.So x(t) is complex signal.

T2 : Solution

(d)(d)(d)(d)(d)Given that x(t) is real

x(t) = 0 for t ≤ 0

and [ ]��

�

� � �

∞ω

−∞

ω ωπ ∫ = �� −

We know that if x(t) ��⎯⎯⎯→ X(ω)

then Even [x(t)],� � � �

�

� �+ −x x ��⎯⎯⎯→ � � � �

�

� � ∗ω + −ω

So Fourier transform of even part of x(t) is real part of X(ω). So inverse Fourier transform of Re[X(ω)] will beeven part of x(t).

So even part of x(t) is �� −

So,� � � �

�

� �+ −x x= �� −

and x(t) = 0 for t < 0So, x(t) = 2te–t u(t)

Fourier Transform3


7Try yourself

T3 : Solution

(a)(a)(a)(a)(a)

We know that if e –2t u(t) ��⎯⎯⎯→�

��ω +

and e2t u(t) ��⎯⎯⎯→ �

� �− ω

T4 : Solution

(a)(a)(a)(a)(a)

Signal g(t) = � ��

��

�∗

πx

Let Fourier transform of x(t) is X(ω).

So Fourier transform of x(t) cos2t will be � � � ��

� � �

ω ω − ω +⎡ ⎤+ +⎢ ⎥⎣ ⎦

Fourier transform of ��

�π is rect(ω/2).

So G(ω) will be � � � ��

��

ω ω − ω +⎡ ⎤+ + ⋅ ω⎢ ⎥⎣ ⎦

Since X(ω) = 0 for �ω >

So G(ω) =� �

�

ω

So, H(ω) = = δ� ��

� ��

T5 : Solution

(c)(c)(c)(c)(c)

Given Fourier transform of e–t u(t) is �

�� + π

Now Fourier transform of �

� �� + π will be e–f u(f ) using the duality theorem.

8 Electrical Engineering • Signals and Systems


T6 : Solution

(a)(a)(a)(a)(a)

Fourier transform is G(ω) = �

�

��

�

ω +ω +

So, G(ω) =�

� �

��

� �

ω +ω + ω +

= �

��

�+

ω +

As we know that Fourier transform of � �

��

� � ��

�

−

+ ω

So g(t) = ( )� � �� δ + −

T7 : Solution

(d)(d)(d)(d)(d)Given that the signal is conjugate symmetric, that is x(t) = x∗(–t)

Let x(t) ��⎯⎯⎯→ X(ω)

then x∗(–t) ��⎯⎯⎯→ X ∗(ω)

Since x(t) = x∗(–t) ⇒ X(ω) = X ∗(ω)So X(ω) is real.

T8 : Solution

(a)(a)(a)(a)(a)

Given that x(t) =� �

��

� �+

Since � ��

− ��⎯⎯⎯→ � �

��

� + ω

then (using duality) � �

��

� �+ ��⎯⎯⎯→ �

��

− ωπ

T9 : Solution

Given that x(t) = 5 Sa(2t) (where Sa is sampling functions)Using the definition of sampling function, we get

x(t) =��

�

�

�

ππ


9Try yourself

The Fourier transform of x(t) will be � ��

ω π

We need to find � � ��

∞

−∞

ω∫

Since, x(t) =�

� ��

� � � �

∞+ ω

−∞

ω ωπ ∫

2π x(0) = � � �

∞

−∞

ω ω∫

So, � � �

∞

−∞

ω ω∫ = 2π x(0) = 31.4


T1 : Solution

(c)(c)(c)(c)(c)To determine the Laplace transform of e–3t cos 2t, we know that

cos2t!��⎯⎯⎯→

� �

�

� +

e–3t cos2t!��⎯⎯⎯→

�

� ��

� ��

�

�

++ +

T2 : Solution

(a)(a)(a)(a)(a)We know that

u(t)!��⎯⎯⎯→ �

�

e–t u(t)!��⎯⎯⎯→

+�

��

e–(t + 1) u(t + 1)!��⎯⎯⎯→

+� ��

��

�

e–t u(t + 1)!��⎯⎯⎯→

+

+

�

� ��

��

�Re{s} > – 1

Laplace Transform4


11Try yourself

T3 : Solution

(a)(a)(a)(a)(a)

Given that Laplace transform of f(t) is ω⎛ ⎞

⎜ ⎟⎝ ⎠+ ω� �� then the signal is sinusoidal and we cannot find the value

of signal at t = ∞.

T4 : Solution

(b)(b)(b)(b)(b)

Given that H(s) =−�

�� "��

� ��

h(t) = (t ∗ e2t) u(t)

T5 : Solution

(d)(d)(d)(d)(d)Given that signal is right sided and system is causal and stable then poles of the system should be in lefthand side of the imaginary axis, so option (d) is correct.


T1 : Solution

To find the value of∞

−

−∞

δ −∫ �� ⇒∞

−

−∞

⎛ ⎞δ −⎜ ⎟⎝ ⎠∫� �

� �

��

=−��

�

�

So, value of integral is 0.303.

T2 : Solution

(a)(a)(a)(a)(a)

Given that y(n) =∞

=−∑ x

� � � #�$�

�

� �

We know that, y(n) =∞

−∞−∑ x � ��

So comparing the above two relation, we geth(n) = 3n u(n)

Discrete Time System5


13Try yourself

T3 : Solution

(a)(a)(a)(a)(a)Given that

x1(n) ⎯⎯⎯⎯→�%� �& y1(n)

δ(n) + δ(n – 1) ⎯⎯⎯⎯→�%� �& cos(n)

and x2(n) ⎯⎯⎯⎯→�%� �& y2(n)

δ(n – 1) + δ(n – 2) ⎯⎯⎯⎯→�%� �& sin(n)

So impulse response can be found as follows:Since x1(n) – x2(n) = δ(n)So impulse response will be y1(n) – y2(n)So, h(n) = cos(n) – sin(n)

T4 : Solution

Given bandwidth (B) of the filter is 3 MHz, lower cut-off frequency is 2 MHz.So upper cut-off frequency (fH) will be 5 MHz.

The sampling frequency (fs) is � �

� (where fH : upper cut-off frequency; k = integer part of fH /B )

So, �

�= =�

��

So, k = 1.0

So, fs = =��'()�

�


T1 : Solution

(b)(b)(b)(b)(b)Let z-transform of x(n) is X(z)

So, x(n) ⎯⎯⎯→*�� X(z)

and ⎛ ⎞⎜ ⎟⎝ ⎠

x� �

�

⎯⎯⎯→*�� X(2z)

Thus,+⎛ ⎞

⎜ ⎟⎝ ⎠x

��

��

⎯⎯⎯→*��

� �

T2 : Solution

(d)(d)(d)(d)(d)Given that x(n) is non zero only for finite values of n, thus ROC of X(z) will be entire z-plane.

T3 : Solution

(d)(d)(d)(d)(d)Given that x(n) = 2n u(n) – 4n u(–n – 1)

So the ROC will be ( ) ( )> ∩ <� ��

Thus, ROC is 2 < ⏐z⏐ < 4.

Z-Transform6


15Try yourself

T4 : Solution

(a)(a)(a)(a)(a)

Given that X(z) =− − − − −= + +

− + + − −� � � � � � �

��

��

= 2(–1)n u(n) + u(2)n u(n) + 12n(2)n u(n)

T5 : Solution

(a)(a)(a)(a)(a)Given that the system has h(n) = 0 for n < 0 and system is not necessarily stable then H(z) should havedegree of numerator less than degree of denominators.Only (I) and (III) satisfy the required condition.

T6 : Solution

(c)(c)(c)(c)(c)Given that x(n) is right sided and real, X(z) has two poles, two zeros at origin and one pole at e jπ/2, X(1) = 1.Since x(n) is real so poles of X(z) should be in conjugate pairs so other pole will be at e–j π/2.

So, X(z) = − π − π =− − +

� �

��

� �

Since, X(1) = 1 so, k = 1

So, X(z) = >+

�

�#�$ �

�

��

�

T7 : Solution

(a)(a)(a)(a)(a)Signal is a nT u(n) then the z-transform will be

−− �

�

� � �=

⎛ ⎞⎜ ⎟⎝ ⎠−

�

� �

T8 : Solution

(b)(b)(b)(b)(b)

Given x(n) = ⎛ ⎞⎜ ⎟⎝ ⎠�

��

�

and − −��

�� = x(n)

So, −− ��

�� = X(z)



⇒ Y(z) =− − − − −

= = − +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

� � � � �

� � � � �

� � � � ��

� � � � �

�

� � � � �

So, y(n) =⎡ ⎤⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

� ��

� �

� �

� � � �


T1 : Solution

(d)(d)(d)(d)(d)

Given, H(s) = + ωω⎛ ⎞+ + ω⎜ ⎟⎝ ⎠

� �

� �

� ��

� ��

So value of H(s) at s → ∞ is kand value of H(s) at s → 0 is k.So the filter is a band stop filter or notch filter.

T2 : Solution

(a)(a)(a)(a)(a)The signal x(t) = (2 + e–3t) u(t) then final value i.e. x(∞) will be 2.

T3 : Solution

(b)(b)(b)(b)(b)Given x(n) signal is real and 8-point DFT is {5, 1 – 3j, 0, 3 – 4j, 3 + 4j, γ, α, β}, we need to find α, βWe know that when signal is real then DFT is conjugate symmetric X(k) = X(–k)∗ and DFT will be periodicwith period 8, that is X(k) = X(8 – k)So α = X(2)∗ = 0

β = X(1)∗ = 1 + 3j

T4 : Solution

(d)(d)(d)(d)(d)Given h(t) = cost u(t)Then system will be causal, unstable and dynamic.

Miscellaneous7



T5 : Solution

(d)(d)(d)(d)(d)Differentiation H(s) = s = jωSo magnitude vary linearly w.r.t. frequency and has constant phase.

T6 : Solution

(c)(c)(c)(c)(c)Properties of distortionless system are:• Magnitude should be constant w.r.t. frequency.• Phase should depend linearly on frequency.Only function given in option (c) follow the given conditions.

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Signals Ty s