SP.1) Which among the following is a linear system ?
(a)$ y[n] = x[n] x[n-1]$
(b)$ y[n] = x[n] + x[n-10]$
(c)$ y[n] = x^2[n]$
(d) both (a) & (c)
Answer is (b)
SP.2) The impulse response of a system is h(t) = t u(t) . For an input u(t-1), the output is
(a)$\frac{t^2 u(t)}{2}$
(b)$\frac{t(t-1)u(t-1)}{2}$
(c)$\frac{(t-1)^2 u(t-1)}{2}$
(d)$\frac{(t^2 - 1)u(t-1)}{2}$
Answer is (c)
SP.3) The impulse response of a continuous time system is given by h(t) = δ(t-1) + δ(t-3); The value of step response at t = 2
(a)0
(b)1
(c)2
(d)3
Answer is (b)
Explanation:
y(t) = h(t)*u(t)
y(t) = [δ(t-1) + δ(t-3)]*u(t)
y(t) = u(t-1)+u(t-3)
y(t) = u(2-1)+u(2-3)
y(t) = u(1) + 0 = 1
SP.4) Consider an LTI system with Transfer function H(s) =1/s(s+4) If the input to the system is cos(3t) and the steady state output is $A sin(3t + \alpha)$ then the value of A is
(a) 1/30
(b) 1/15
(c) 3/4
(d) 4/3
Answer is (b)
Explanation:
$ H(s) = \frac{1}{s(s+4)} $
$ X(s) = \frac{s}{s^2+9} $
$ Y(s) = A\frac{s sin(\alpha) + 3cos(\alpha)}{s^2+9}$
$|Y(s)|=|H(s)||X(s)|$
$cos(\omega t) = cos(3t)$
$\omega = 3$
A = 1 . $\frac{1}{\omega\sqrt{\omega^2+16}}$
SP.5) For the discrete time system of the given figure:
(a) y(k) - 0.5 y(k-1) - 0.25 y(k-2) = u(k)
(b) y(k) - 0.5 y(k-1) + 0.25 y(k-2) = u(k)
(c) y(k) + 0.5 y(k-1) - 0.25 y(k-2) = u(k)
(d) y(k) + 0.5 y(k-1) + 0.25 y(k-2) = u(k)
Answer is (b) Explanation
From figure,
y(k) = u(k) + 0.5 y(k-1) - 0.25 y(k-2)
Thus, u(k) = y(k) - 0.5 y(k-1) + 0.25 y(k-2)
SP.6)ROC of the sequence x[n] = a^n u[n] is
(a) z > a
(b) z < a
(c) |z| > a
(d) |z| < a
SP.7) For the analog signal m(t) = 4 cos(100πt) + 8 sin(200πt) + cos(300πt) , the Nyquist Sampling Rate will be
(a) 1/100
(b) 1/200
(c) 1/300
(d) 1/600
Explanation
Nyquist Sampling frequency must be the twice the highest message frequency in the signal.
Further, Nyquist Sampling rate is the reciprocal of Nyquist Sampling frequency.
Thus Nyquist sampling rate = 1/(2*150) = 1/300. Here 150 is the highest frequency of the signal m(t) [as evident from cos(300πt) when expressed as cos(2*150*πt).]
SP.8) In Laplace transform, multiplication by e^(-at) in time domain becomes
(a) translation by a in s-domain
(b) translation by (-a) in s-domain
(c) multiplication by e^(-as) in s-domain
(d) none of above
SP.9) Inverse Fourier transform of sgn(ω)is
(a) j/πt
(b) 1
(c) u(t)
(d) 2/jt
SP.10) For an input x(t) , the output y(t) of a modulator is given by
y(t) = x(t) cos(2πft) where f(t) is carrier frequency. The system is
(a) dynamic,nonlinear & time invariant
(b) static,linear & time-variant
(c) static,non-linear & time-variant
(d) dynamic,linear & time-variant
SP.11) A signal m(t) is multiplied by a sinusoidal waveform of a frequency fc such that v(t) = m(t)cos(2π fc t). If Fourier Transform of m(t) is M(f), Fourier Transform of v(t) will be
(a) 0.5 M(f + fc)
(b) 0.5 M(f - fc)
(c) 0.5 M(f + fc) + 0.5 M(f - fc)
(d) 0.5 M(f - fc) + 0.5 M(f - fc)
Answer is (c)
SP.12) For a periodic signal v(t) = 30 sin (100t) + 10 cos 300t + 6 sin (500+π/4), the fundamental frequency in rad/s
(a) 100
(b) 300
(c) 500
(d) 1500
Answer is (a)
SP.13) For the function x(t) shown in figure
even & odd parts of a unit step function u(t) are respectively
(a) 0.5,0.5 x(t)
(b) -0.5,0.5 x(t)
(c) 0.5,-0.5 x(t)
(d) -0.5,-0.5 x(t)
Answer is (a)
SP.14) The following is true
(a) A Finite Signal is Always Bounded
(b) A Bounded Signal Always possess Finite Energy
(c) A Bounded Signal is Zero outside the interval [-t0,t0] for some t0
(d) A Bounded Signal is Always Finite
Answer is (b)
(a)$ y[n] = x[n] x[n-1]$
(b)$ y[n] = x[n] + x[n-10]$
(c)$ y[n] = x^2[n]$
(d) both (a) & (c)
Answer is (b)
SP.2) The impulse response of a system is h(t) = t u(t) . For an input u(t-1), the output is
(a)$\frac{t^2 u(t)}{2}$
(b)$\frac{t(t-1)u(t-1)}{2}$
(c)$\frac{(t-1)^2 u(t-1)}{2}$
(d)$\frac{(t^2 - 1)u(t-1)}{2}$
Answer is (c)
SP.3) The impulse response of a continuous time system is given by h(t) = δ(t-1) + δ(t-3); The value of step response at t = 2
(a)0
(b)1
(c)2
(d)3
Answer is (b)
Explanation:
y(t) = h(t)*u(t)
y(t) = [δ(t-1) + δ(t-3)]*u(t)
y(t) = u(t-1)+u(t-3)
y(t) = u(2-1)+u(2-3)
y(t) = u(1) + 0 = 1
SP.4) Consider an LTI system with Transfer function H(s) =1/s(s+4) If the input to the system is cos(3t) and the steady state output is $A sin(3t + \alpha)$ then the value of A is
(a) 1/30
(b) 1/15
(c) 3/4
(d) 4/3
Answer is (b)
Explanation:
$ H(s) = \frac{1}{s(s+4)} $
$ X(s) = \frac{s}{s^2+9} $
$ Y(s) = A\frac{s sin(\alpha) + 3cos(\alpha)}{s^2+9}$
$|Y(s)|=|H(s)||X(s)|$
$cos(\omega t) = cos(3t)$
$\omega = 3$
A = 1 . $\frac{1}{\omega\sqrt{\omega^2+16}}$
SP.5) For the discrete time system of the given figure:
(a) y(k) - 0.5 y(k-1) - 0.25 y(k-2) = u(k)(b) y(k) - 0.5 y(k-1) + 0.25 y(k-2) = u(k)
(c) y(k) + 0.5 y(k-1) - 0.25 y(k-2) = u(k)
(d) y(k) + 0.5 y(k-1) + 0.25 y(k-2) = u(k)
Answer is (b) Explanation
From figure,
y(k) = u(k) + 0.5 y(k-1) - 0.25 y(k-2)
Thus, u(k) = y(k) - 0.5 y(k-1) + 0.25 y(k-2)
SP.6)ROC of the sequence x[n] = a^n u[n] is
(a) z > a
(b) z < a
(c) |z| > a
(d) |z| < a
SP.7) For the analog signal m(t) = 4 cos(100πt) + 8 sin(200πt) + cos(300πt) , the Nyquist Sampling Rate will be
(a) 1/100
(b) 1/200
(c) 1/300
(d) 1/600
Explanation
Nyquist Sampling frequency must be the twice the highest message frequency in the signal.
Further, Nyquist Sampling rate is the reciprocal of Nyquist Sampling frequency.
Thus Nyquist sampling rate = 1/(2*150) = 1/300. Here 150 is the highest frequency of the signal m(t) [as evident from cos(300πt) when expressed as cos(2*150*πt).]
(a) translation by a in s-domain
(b) translation by (-a) in s-domain
(c) multiplication by e^(-as) in s-domain
(d) none of above
SP.9) Inverse Fourier transform of sgn(ω)is
(a) j/πt
(b) 1
(c) u(t)
(d) 2/jt
SP.10) For an input x(t) , the output y(t) of a modulator is given by
y(t) = x(t) cos(2πft) where f(t) is carrier frequency. The system is
(a) dynamic,nonlinear & time invariant
(b) static,linear & time-variant
(c) static,non-linear & time-variant
(d) dynamic,linear & time-variant
SP.11) A signal m(t) is multiplied by a sinusoidal waveform of a frequency fc such that v(t) = m(t)cos(2π fc t). If Fourier Transform of m(t) is M(f), Fourier Transform of v(t) will be
(a) 0.5 M(f + fc)
(b) 0.5 M(f - fc)
(c) 0.5 M(f + fc) + 0.5 M(f - fc)
(d) 0.5 M(f - fc) + 0.5 M(f - fc)
Answer is (c)
SP.12) For a periodic signal v(t) = 30 sin (100t) + 10 cos 300t + 6 sin (500+π/4), the fundamental frequency in rad/s
(a) 100
(b) 300
(c) 500
(d) 1500
Answer is (a)
SP.13) For the function x(t) shown in figure
even & odd parts of a unit step function u(t) are respectively
(a) 0.5,0.5 x(t)
(b) -0.5,0.5 x(t)
(c) 0.5,-0.5 x(t)
(d) -0.5,-0.5 x(t)
Answer is (a)
SP.14) The following is true
(a) A Finite Signal is Always Bounded
(b) A Bounded Signal Always possess Finite Energy
(c) A Bounded Signal is Zero outside the interval [-t0,t0] for some t0
(d) A Bounded Signal is Always Finite
Answer is (b)
