Linear System Signal Analysis Practice

The signal is continuous-time because it is defined for real values of $t$.

It is deterministic because it is given by an explicit formula, not a statistical description.

Replace $t$ by $t-3$:

This is the original step delayed by $3$ units.

First compute $x(-t)$:

Now use the formulas:

So

and

No. Linearity requires superposition.

For an input $x(t)$, the output is $x(t)^2$. But if we double the input, the output becomes

whereas doubling the original output gives

These are not the same, so the system is not linear.

The system has memory because $y(t)$ depends on $x(t-1)$, not just the input at the same time.

It is causal because it depends only on the present value $x(t)$ and the past value $x(t-1)$, not on any future input.

An impulse delayed by $2$ units produces a delayed copy of the impulse response:

This follows from the shift property of convolution.

Convolution with a delayed impulse shifts the other signal:

So

Use

Substitute $T = 0.25$:

So the fundamental angular frequency is

The pole is where the denominator is zero:

Since $-5$ has negative real part, the pole lies in the left-half plane.

Compare the equations to

and

So

The system is linear because $x(t)$ and $y(t)$ appear only to the first power and are not multiplied together.

It is time invariant because the coefficients are constant and there is no explicit dependence on $t$.

So the system is LTI.

First compute

Then

and

So

and

The filter averages three consecutive samples:

Since $x[n]$ is a length-3 pulse, the output is

For a cosine input at frequency $\omega$, the output keeps the same frequency and is scaled and phase shifted by $H(j\omega)$.

The magnitude is $2$ and the phase is $-\pi/6$, so

Use the time-shift property:

With $t_0 = 3$,

So

Take the Laplace transform of both sides:

So

Use partial fractions:

Invert term by term:

Take the Z-transform with zero initial conditions:

Factor out $Y(z)$:

So

For the causal system, the ROC is

The Nyquist rate is

So the sampling rate must be greater than $1800$ Hz to avoid aliasing.

Since $1.5$ kHz $= 1500$ Hz is below $1800$ Hz, it is not sufficient.

Use linearity and the shift property:

Substitute $h(t)$:

Evaluate the frequency response at $\omega=2$:

Its magnitude is

and its phase is

So the output amplitude is

and the steady-state output is

The sampled spectrum repeats every $f_s = 4$ kHz.

Since

the tone folds to

in the baseband interval $[0, f_s/2]$.

Use

with

Then

So

Odd symmetry means the Fourier series can contain sine terms only. There is no DC term and no cosine terms.

The fundamental frequency is

So the harmonics occur at integer multiples of $100$ Hz.

Both signals are rectangular pulses. The convolution equals the amount of overlap between the interval $[0,2)$ and the shifted interval $[t-1,t)$.

That overlap length is:

So the output is a trapezoid made of two ramps and a flat middle section.

Take the Laplace transform:

So

Use partial fractions:

Invert term by term:

The poles are at

For a causal continuous-time rational system, the ROC must lie to the right of the rightmost pole, so here the ROC would be

That ROC does not include the $j\omega$ axis, so the system would not be stable.

Therefore, a causal stable realization is not possible.

The frequency response is

so its magnitude is

and its phase is

Therefore the steady-state output is