List of top Signals and Systems Questions

A counter uses three D flip-flops generating the Gray code sequence 000, 001, 011, 010, 110, 111, 101, 100 repeatedly. The bits are in $Q_2 Q_1 Q_0$ format. The combinational logic expression for $D_1$ is $\underline{\hspace{1cm}}$.

A 16-bit synchronous binary up-counter is clocked with frequency $f_{CLK}$. The two most significant bits are OR-ed to form output Y. Y remains high for 24 ms. The clock frequency $f_{CLK}$ is $\underline{\hspace{1cm}}$ MHz. (Round off to 2 decimal places.)

Two discrete-time linear time-invariant systems with impulse responses $h_1[n] = \delta[n-1] + \delta[n+1]$ and $h_2[n] = \delta[n] + \delta[n-1]$ are connected in cascade. The impulse response of the cascaded system is

When acceptance number of a single sampling plan under attribute category is zero with sample size less than or equal to 10, the Operating Characteristic (OC) curve is

Consider a carrier signal which is amplitude modulated by a single-tone sinusoidal message signal with a modulation index of 50%. If the carrier and one of the sidebands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is .

In the circuit shown in the figure, the transistors M1 and M2 are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the MOSFETs M1 and M2 are $ g_{m1} $ and $ g_{m2} $, respectively, and the internal resistance of the MOSFETs M1 and M2 are $ r_{o1} $ and $ r_{o2} $, respectively. Ignoring the body effect, the ac small signal voltage gain $ \frac{\partial V_{\text{out}}}{\partial V_{\text{in}}} $ of the circuit is

For a unit step input $ u[n] $, a discrete-time LTI system produces an output signal $ \left( 2\delta[n + 1] + \delta[n] + \delta[n - 1] \right) $. Let $ y[n] $ be the output of the system for an input $ \left(\frac{1}{2}\right)^n u[n] $. The value of $ y[0] $ is _________.

A speech signal, band limited to 4 kHz, is sampled at 1.25 times the Nyquist rate. The speech samples, assumed to be statistically independent and uniformly distributed in the range $ -5 \, \text{V} $ to $ +5 \, \text{V} $, are subsequently quantized in an 8-bit uniform quantizer and then transmitted over a voice-grade AWGN telephone channel. If the ratio of transmitted signal power to channel noise power is 26 dB, the minimum channel bandwidth required to ensure reliable transmission of the signal with arbitrarily small probability of transmission error (rounded off to two decimal places) is kHz.

A 4 kHz sinusoidal message signal having amplitude 4 V is fed to a delta modulator (DM) operating at a sampling rate of 32 kHz. The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is V.

Consider two 16-point sequences $x[n]$ and $h[n]$. Let the linear convolution of $x[n]$ and $h[n]$ be denoted by $y[n]$, while $z[n]$ denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of $x[n]$ and $h[n]$. The value(s) of $k$ for which $z[k] = y[k]$ is/are

Consider a real-valued base-band signal $ x(t) $, band limited to 10 kHz. The Nyquist rate for the signal $ y(t) = x(t) \cdot \left( 1 + \frac{t}{2} \right) $ is

In the block diagram shown below, an infinite tap FIR filter with transfer function $ H(z) = \frac{Y(z)}{X(z)} $ is realized. If \[ H(z) = \frac{1}{1 - 0.5z^{-1}}, \] the value of $ \alpha $ is $\underline{\hspace{2cm}}$

$x[n]$ is convolved with $h[n]$ to give $y[n]$. If $y[2] = 1$ and $y[3] = 0$, then $h[0] = \underline{\hspace{1cm}}.$ (Graphs are not uniformly scaled)

A unit step input is applied to a system with impulse response $ H(s) = \dfrac{1-s/\omega_z}{1+s/\omega_p} $ at $ t = 0 $. The output of the system $ y(t) $ at $ t = 0^+ $ is

A continuous time transfer function $ H(s) = \dfrac{1 + s/10^6}{s} $ is converted to a discrete time transfer function $ H(z) $ using a bilinear transform at a sampling rate of 100 MHz. The pole of $ H(z) $ is located at $ z = $ $\underline{\hspace{2cm}}$.

Let $X(j\omega)$ denote the Fourier transform of $x(t)$. If $X(j\omega) = 10 e^{-j\pi f} \left( \frac{\sin(\pi f)}{\pi f} \right)$, then $ \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) d\omega = \underline{\hspace{1cm}}. $ (where $\omega = 2\pi f$)

An analog signal is sampled at 100 MHz to generate 1024 samples. Only these samples are used to evaluate 1024-point FFT. The separation between adjacent frequency points ($ \Delta F $) in FFT is $\underline{\hspace{2cm}}$ kHz.

The sinusoidal T.F. Corresponding to the polar plot for T>0.