Signal Processing MCQ Questions & Answers | ECE

The impulse response of a linear time invariant system is h(n) = {1, \[\underline 2 \], 1, -1}. The response for the input signal x(n) = {\[\underline 1 \], 2, 3, 1} is

With the following equations, the time invariant systems are
$$\eqalign{ & 1.\,\frac{{{d^2}y\left( t \right)}}{{d{t^2}}} + 2t\frac{d}{{dt}}y\left( t \right) + 5y\left( t \right) = x\left( t \right) \cr & 2.\,y\left( t \right) = {e^{ - 2x\left( t \right)}} \cr & 3.\,y\left( t \right) = \left[ {\frac{d}{{dt}}x\left( t \right)} \right] \cr & 4.\,y\left( t \right) = \frac{d}{{dt}}\left[ {{e^{ - 2t}}x\left( t \right)} \right] \cr} $$

Two ideal quantizers A and B have following specifications:
A: 5 bit Quantizer with input dynamic range of -1V to +1V with Q₁ as quantization noise power
B: 8 bit Quantizer with input dynamic range of -0.5V to +0.5V with Q₂ as quantization noise power.
Then $$\frac{{{{\text{Q}}_1}}}{{{{\text{Q}}_2}}}$$ will be

The trigonometric Fourier series of an even function of time does not have

Laplace transform of 1 - e^-at is given by

The dead band range of a first order IIR filter given by y(n) = 0.5y (n - 1) + x(n) on quantization operation due to rounding with a step size of is:

The filter used for pulse shaping is

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 $$y\left( t \right) = \int\limits_{ - \infty }^t {g\left( \tau \right)d\tau ,} $$    then $$\int\limits_{ - \infty }^{ + \infty } {y\left( t \right)dt} $$   is . . . . . . . .

An LTI system has a wide-sense stationary (WSS) input signal with zero mean, Its output is

The Dirac delta function δ(t) is defined as