GATE Electrical Engineering (EE) Test: Introduction to Systems Free Online

For even function, f(t) = f(-t)

For odd function, f(t) = -f(-t)

Convolution in the time domain implies multiplication in S(or frequency) domain

The Laplace transform of any signal h(t) is given by

So we observe that the H(s) = 1/s³, corresponds to a unit step signal convoluted with itself thrice.

Therefore the correct answer is option 1

Given: h(t) = u(t)
x(t) = e^–at u(t)

∴ Y(s) = X(s) H(s) =   

y(t) = 1/a (1 – e^{– at}) u(t)

y(t) = tx(t)

y₁(t) = t.x₁(t) = r₁(t)

y₂(t) = tx₂(t) = r₂(t)

y₁(t) + y₂(t) = t(x₁(t) + x₂(t))

= r₁(t) + r₂(t)    ∴ linear

y(t) = t.x(t)

y( t - t_o) = (t - t_o) x ( t - t_o)

and for delayed input signal,

y(t) = t x (t - t_o)

y(t) ≠ y( t-t_o)

∴ Time varying signal

Since f(t) u(t) = f(t) for t > 0 also we know

u ( t - t_o) = 1, for t > t_o

Here in right side shifting that means t_o > 0

by property on shifting right side,

Since, causal system

h(t) = 0 for t < 0

Input y(t) = x(t) * h(t)

h(τ) = 0 for τ < 0

x ( t - τ) = 0, for t - τ < - 2

∴ τ > f + 2

Given x(t) and h(t) y(t) = x(t) * h(t)

y(t) =^∞x(τ) h(t – τ)dτ_{– ∞}

Thus, we conclude that the output of the system is zero everywhere except for the time interval.
1 < t < 5

The problem involves a signal f(t) = cos8πt + 0.5cos4πt. To find the maximum allowable sampling interval, we utilise the Nyquist-Shannon sampling theorem, which states that the sampling rate must be at least twice the highest frequency present in the signal.

• The signal consists of two cosine terms: cos8πt and 0.5cos4πt.
• The frequencies of these terms are 4 Hz and 2 Hz respectively.
• The highest frequency component is 4 Hz.
• According to the sampling theorem, the minimum sampling rate is 2 × 4 Hz = 8 Hz.
• The maximum allowable sampling interval T_s is the reciprocal of the sampling rate, thus 1/8 sec.