Signal processing

1. {a(n)} is a real-valued periodic sequence with a period N. x(n) and X(k) form N-point Discrete Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence $$y\left( n \right) = \frac{1}{N}\sum\limits_{r = 0}^{N – 1} {x\left( r \right)} x\left( {n + r} \right)$$ is

$${left| {Xleft( k ight)} ight|^2}$$
$$rac{1}{2}sumlimits_{r = 0}^{N - 1} {Xleft( r ight)X'left( {k + r} ight)} $$
$$rac{1}{2}sumlimits_{r = 0}^{N - 1} {Xleft( r ight)Xleft( {k + r} ight)} $$
0

Detailed Solution{a(n)} is a real-valued periodic sequence with a period N. x(n) and X(k) form N-point Discrete Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence $$y\left( n \right) = \frac{1}{N}\sum\limits_{r = 0}^{N – 1} {x\left( r \right)} x\left( {n + r} \right)$$ is

2. For a function g(t), it is given that $$\int\limits_{ – \infty }^{ + \infty } {g\left( t \right){e^{ – j\omega t}}dt = \omega {e^{ – 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ – \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ – \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

0
#NAME?
$$ - {j over 2}$$
$${j over 2}$$

Detailed SolutionFor a function g(t), it is given that $$\int\limits_{ – \infty }^{ + \infty } {g\left( t \right){e^{ – j\omega t}}dt = \omega {e^{ – 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ – \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ – \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

4. Let Y(s) be the unit-step response of a causal system having a transfer function $$G\left( s \right) = {{3 – s} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ That is, $$Y\left( s \right) = {{G\left( s \right)} \over s}.$$ The forced response of the system is

u(t) - 2e-t u(t) + e-3t u(t)
2u(t)
u(t)
2u(t) - 2e-t u(t) + e-3t u(t)

Detailed SolutionLet Y(s) be the unit-step response of a causal system having a transfer function $$G\left( s \right) = {{3 – s} \over {\left( {s + 1} \right)\left( {s + 3} \right)}}$$ That is, $$Y\left( s \right) = {{G\left( s \right)} \over s}.$$ The forced response of the system is

5. The Laplace transform of a continuous-time signal x(t) is $$X\left( s \right) = {{5 – s} \over {{s^2} – s – 2}}.$$ If the Fourier transform of this signal exists, then x(t) is

e2tu(t) - 2e-tu(t)
-e2tu(-t) + 2e-tu(t)
-e2tu(-t) - 2e-tu(t)
e2tu(-t) - 2e-tu(t)

Detailed SolutionThe Laplace transform of a continuous-time signal x(t) is $$X\left( s \right) = {{5 – s} \over {{s^2} – s – 2}}.$$ If the Fourier transform of this signal exists, then x(t) is

9. A real-valued signal x(t) limited to the frequency band $$\left| f \right| \le {W \over 2}$$ is passed through a linear time invariant system whose frequency response is $$H\left( f \right) = \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is

x(t + 4)
x(t - 4)
x(t + 2)
x(t - 2)

Detailed SolutionA real-valued signal x(t) limited to the frequency band $$\left| f \right| \le {W \over 2}$$ is passed through a linear time invariant system whose frequency response is $$H\left( f \right) = \left\{ {\matrix{ {{e^{ – j4\pi f,}}} & {\left| f \right| \le {W \over 2}} \cr {0,} & {\left| f \right| > {W \over 2}} \cr } } \right.$$ The output of the system is

10. The z-transform of a system is $$H\left( z \right) = {z \over {z – 0.2}}.$$ If the ROC is |z| < 0.2, then the impulse response of the system is

”(0.2)nu[n
” option2=”(0.2)nu[- n – 1]” option3=”-(0.2)nu[n]” option4=”-(0.2)nu[- n – 1]” correct=”option3″]

Detailed SolutionThe z-transform of a system is $$H\left( z \right) = {z \over {z – 0.2}}.$$ If the ROC is |z| < 0.2, then the impulse response of the system is

Page 1Page 2Page 3Page 4Page 5
Test 1Test 2Test 3
Exit mobile version