All India Electronics and Communication Engineering (ECE) Group

In time reversal only the time variable is affected and not the amplitude.

In time scaling the signal is compressed or expanded by multiplying the time variable

When the α value is less than one in time scaling then the signal is said to be expanded or dilated

Amplitude scaling is represented with the equation Y(t) = α X(t)

n time reversal we multiply the time variable by -1.

In time signal finite time value is added or subtracted from it.

The given y (t) is integral of x (t) and amplitude 3 remains constant for t > 1.
It is because of the properties of integration.

For discrete time signal y [n] = x [k x n] and k > 1, it will be compressed signal and some samples will be lost. The samples lost will not violate the rules of sampling theorem.

By comparing the given equation with y (t) = x (at) we get a = 2. If a > 1 then it is compressed version of x (t).

Capacitor performs integration. V (t) developed across capacitor is given by
v (t) = (1/C)* ∫^t_-∞ i (∂).d∂, I (t) is the current flowing through a capacitor of capacitance C.

y (t) = x (at), comparing this with the given expression we get a = 1/5. If 0 < a < 1 then it is expanded (stretched) version of x (t).

Time scaling is an example for operations performed on independent variable time.
It is given by y (t) = x (at).

Inductor performs differentiation. It is given by y (t) = L d/dt i(t) where, I (t) denotes current flowing through an inductor of inductance L.

Audio mixer is a device which combines music and voice signals. It is given by
Y (t) = x₁ (t) + x₂ (t).

AM radio signal is an example for y (t) = x₁ (t) * x₂ (t) where, x₁ (t) consists of an audio signal plus a dc component and x₂ (t) is a sinusoidal signal called carrier wave.

Amplitude scaling refers to multiplication of a constant with the given signal.
It is given by y (t) = a x (t). It can be both increase in amplitude or decrease in amplitude.

We know that Nyquist frequency is twice the maximum frequency, i.e. f_s = 2 f_m.
The maximum frequency present in the signal is ω_m = 300 π or f_m = 150 Hz. Therefore the Nyquist frequency f_s = 2 f_m = 300 Hz.

DTF System operates with a discrete signal, on the other hand time invariant system is a system whose output does not depend explicitly on time. For continuous time system, the inputs as well as output both are CT signals.

Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0
For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)
∴ y(0) = x(0) – x(-1)
Or, y (n) = 0 for n<0
For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)
Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)
Or, y (1) = x (1) +x (0) – 2x (-1)
For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)
Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)
∴y (2) = d (2) + d (1) + d (0) – 3d (-1).

Given that, N₁ = 4, N₂ = 16, N₃ = 8
We know that period of X [n] (say N) = LCM (N₁, N₂, N₃)
∴ Period of X [n] = LCM (4, 16, 8) = 16.

The steady state value of the DT signal F(s) exists since all poles of the given Laplace transform have negative real part.
∴F (∞) = lim_s→0 s F(s)

= 0.

We know that for X [n] = X₁ [n] × X₂ [n] to be periodic, both X₁ [n] and X₂ [n] should be periodic with finite periods.
Here X₂ [n] = cos (πn/8), is periodic with fundamental period as 8/n
But X1 [n] = cos (n/8) is non periodic.
∴ X [n] is a non-periodic signal.

Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z^-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z^-n.

We know that any signal be it discrete or continuous is said to be even or symmetric when that signal f(x) = f (-x). Here given signal is X (n). It is a discrete time signal. So, the signal will be even symmetric if X (n) = X (-n).

Given that, N₁ = 18, N₂ = 14
We know that period of X [n] (say N) = LCM (N₁, N₂)
∴ Period of X [n] = LCM (18, 14) = 126.

y₁ (t) = v (t - 5) – v (3 - t)
y₂ (t) = k v (t - 5) – k v (3 - t) = ky₁ (t)
Let x₁ (t) = v (t), then y₁ (t) = v (t - 5) – v (3 - t)
Let x₂ (t) = 2w (t), then y2 (t) = w (t - 5)-w (3 - t)
Let x₃ (t) = x (t) + w (t)
Then, y₃ (t) = y1 (t) + y₂ (t)
Hence it is linear.
Again, y₁ (t) = v (t - 5) – v (3 - t)
∴ y₂ (t) = y1 (t-t0)
Hence, system is time-invariant
If x (t) is bounded, then, x (t-5) and x (3-t) are also bounded, so stable system.
At t = 0, y (0) = x (-5)-x (3), therefore, the response at t=0 depends on the excitation at a later time t = 3.
Therefore Non-Causal.

y[k] = x [3k-2]
Now, y [0] = x [-2] = 1
Or, y [1] = x [1] = 1
Or, y [2] = x [4] = 0
∴ y[k] = 1, for k = 0, 1 and 0 otherwise.

⇒ {—5, 6, 17, 42, | 38, 30, 8}

= 

Hence, option B is correct.

Also,

Hence, option C is also correct.

Apply ‘z’ transform on both sides.

y(n) + 5y(n-1) + 6y(n-2) = 5x(n) – 2x(n-1)

Taking the z-transform

Y(z) + 5z^-1 Y(z) + 6z^-2 Y(z) = 5X(z) - 2z^-1X(z)

y(z) [1 + 5z^-1 + 6z^-2] = X(z) [5 – 2z^-1]

Taking inverse Z-transform,

h(n) = 17(-3)ⁿ u(n) – 12(-2)ⁿ u(n)

For an LTI system the output is given by

y(n) = x(n) ⨂ [h₁(n) ⨂ h₂(n)]

given

h₁(n) = δ(n)-5δ(n­-1)

h₂(n) = 5ⁿ u(n)

h₁(n) ⨂ h₂(n) = [δ(n) - 5δ(n-1) ⨂ 5ⁿu(n)

= 5ⁿu(n) – 5 (5^n-1 u(n-1))

= 5ⁿu(n) – 5ⁿu(n-1)

= 5ⁿ [u(n) – u(n-1)]

A = -5

B = 9

Poles of the H(z) is at |z| = 2, |z| = 3. So, these are 3 possible ROC.

Case (1) 2 < |z|<3

Case (2) |z| ≤ 2

Case (3) |z| ≥ 3

Only in case (2) signal is anti-causal and stable for stable ROC must cuts the unity circle so for ROC |z| ≤ 2

h(n)=-5(-2ⁿu(-n-1) + 9(-3ⁿu(-n-1)

h(n)=5×2ⁿu(-n-1) – 9 × 3ⁿu(-n-1)

Note: u(n) – u(n-1) = δ(n)

so h₁(n) ⨂ h₂(n) = 5ⁿδ(n) = δ(n)

y(n) = x(n) ⨂ [h₁(n) ⨂ h₂(n)]

= (1/3)ⁿu(n) ⨂ [δ(n)]

We have x′(t)=

x′(t) = r(t) – 2r(t − 2) + r(t − 4)

Now,x(t) = x′(2t) compression by 2.

Thus,

x(t) = r(2t) – 2r(2t − 2) + r(2t − 4).

Alternatively, x(t) may also be written as

Concept:

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left. 

For example:

Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'. 

If a > 1, then the signal is contracted by a factor of 'a' along the time axis. 

if a < 1, then the signal is expanded by a factor of 'a' along the time axis. 

For example:

We can observe that if we scale f(t) by a factor of ½ and then shift, we will get g(t).

First scale f(t) by a factor of ½ ,

g₁(t) = f (t/2)

Shift g₁(t) by 3

 

z₁[n] = x[n - 3]

z₂[n] = z₁[4n] = x[4n - 3]

y[n] = z₂[-n] = x[-4n - 3] ≠ x[3 - 4n]

z₁[n] = x[n + 3]

z₂[n] =z₁[4n] = x[4n + 3]

y[n] = z₂[-n] = x[-4n + 3]

v₁[n] = x[4n]

v₂[n] = v₁[-n] = x[-4n]

y[n] = v₂[n+3] = x[-4(n + 3)] ≠ x[3 - 4n]

v₁[n] = x[n]

x₂[n] = v₁[-n] = x[-4n]

y[n] = v₂[n-3] = x[-4(n - 3)] ≠ x[3 - 4n]

Given x(t) = u(sint)
Drawing waveform

We know for waveform like y(t) shown

DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics

We know that,

Given form is,

So, we can write it as,

Comparing both equations, we get,

Taking Inverse Fourier Transform,

Comparing with given signal, g(t) = Ay (Bt)
A = 1/3 and B = 3

We know that from Fourier Series coefficient of square periodic signal, power is proportional to (1/n²).

CASE 1: Let us assume that x(t) is real, then x(t) = x*(t). This implies that for x(t) to be real, ak = a*_-k. Since this is not true. So, x(t) is not real.
CASE 2: Let us assume that x(t) is even, then x(t) = x(-t) and a_k = a_–k. Since this is true for this case, hence x(t) is even.
CASE 3: Let us assume

Therefore,

Since bk is not even, So, g(t) is not even.

The average value