Unit Impulse/Delta Signal MCQ Quiz - Objective Question with Answer for Unit Impulse/Delta Signal - Download Free PDF

$\mathrm{y}(\mathrm{n})=\sum _{\mathrm{n}=-5}^5\sin (2\mathrm{n})\delta (\mathrm{n}+7)$

n + 7 = 0

n = -7

which does not lie in between -5 ≤ n ≤ 5

hence y[5] = 0

Concept:

$\delta \left[at+b\right]=\frac{1}{\left|a\right|}\delta \left[t+\frac{b}{a}\right]$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right).\delta \left(t-a\right)=f\left(a\right)$

Calculation:

$\delta \left(3t-\pi \right)=\frac{1}{\left|3\right|}\delta \left[t-\frac{\pi }{3}\right]$

$=\frac{1}{3}\delta \left[t-\frac{\pi }{3}\right]$

Given,

$\therefore {\int }_{\frac{-\pi }{6}}^{\frac{\pi }{6}}\sin \left(3t-\frac{\pi }{2}\right).\delta \left(3t-\pi \right).dt$

$={\int }_{\frac{-\pi }{6}}^{\frac{\pi }{6}}\sin \left(3t-\frac{\pi }{2}\right).\frac{1}{3}\delta \left(t-\frac{\pi }{3}\right).dt$

$=\frac{1}{3}{\int }_{\frac{-\pi }{6}}^{\frac{\pi }{6}}\sin \left(3t-\frac{\pi }{2}\right).\delta \left(t-\frac{\pi }{3}\right).dt$

= 0

Given $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)\delta (t)dt$,

x(t) = u(t - 2) - u (t - 4)

x(t) =1 ,  when t lies in between 2 and 4

x(t)= 0, in all other cases

The sifting property of the impulse (delta) function is defined as :

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)\delta (t)dt$ = x(0) = 0

The answer is zero.

Option 2 is correct.

$\mathrm{y}(\mathrm{n})=\sum _{\mathrm{n}=-5}^5\sin (2\mathrm{n})\delta (\mathrm{n}+7)$

n + 7 = 0

n = -7

which does not lie in between -5 ≤ n ≤ 5

hence y[5] = 0

Concept:

Unit Impulse function:

A continuous-time unit impulse function δ(t), also called a Dirac delta function is defined as:

$\delta \left(t\right)=\{\begin{array}{c}\mathrm{\infty },t=0\\ 0,t\ne 0\end{array}$

The unit impulse function is represented by an arrow with the strength of  ‘1’ which represents its area.

Area =

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

Properties of δ(t):

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0)δ(t – t0) 

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$

5. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt=(-1{)}^n{\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

Even Function : x(t) = x(-t)

Odd Function: x(t) = - x(-t)

Explanation:

δ (t) is an even function from its definition, so δ(t) = δ(- t)

From property, 

$\delta (2t)=\frac{1}{|2|}\delta (t)=\frac{1}{2}\delta (t)$

Important Points.

Properties of δ(n):

1. x[n] δ[n] = x[0] δ[n]

2. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[0\right]\delta \left[n\right]=x\left[0\right]$

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n-n_0\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n_0\right]\delta \left[n-n_0\right]=x\left[n_0\right]$

5. δ[an] = δ[n]

Concept:

Shifting property of impulse function

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{x}(\mathrm{t})\delta (\mathrm{t}-\mathrm{a})\mathrm{d}\mathrm{t}=\mathrm{x}(\mathrm{a})}$

Scaling property of impulse function

$\delta (at)=\frac{1}{|a|)}\delta (t)$

Explanation:

Let:

$\mathrm{I}={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\delta (2\mathrm{t}-2)\mathrm{d}\mathrm{t}}$

$I={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\delta [2(\mathrm{t}-1)]\mathrm{d}\mathrm{t}}$

Using scaling property of impulse function in the above equation, we'll get:

$I={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\frac{1}{|2|}\delta (\mathrm{t}-1)\mathrm{d}\mathrm{t}}$

Applying Shifting property of impulse function to the above equation, we'll get:

$I=\frac{1}{2}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\delta (\mathrm{t}-1)\mathrm{d}\mathrm{t}}$

$I=\frac{1}{2}.{e^{-t}|}_{t=1}$

$\frac{1}{2e}$

Unit impulse function:

It is defined as

Properties:

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0)

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$

5. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt={\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

The discrete time version of the unit impulse is defined by

$\delta \left[n\right]=\{\begin{array}{c}1,n=0\\ 0,n\ne 0\end{array}$

Properties:

1. x[n] δ[n] = x[0] δ[n]

2. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[0\right]\delta \left[n\right]=x\left[0\right]$

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n-n_0\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n_0\right]\delta \left[n-n_0\right]=x\left[n_0\right]$

5. δ[an] = δ[n]

Concept:

Unit impulse function:

It is defined as, $\delta \left(t\right)=\{\begin{array}{c}\mathrm{\infty },t=0\\ 0,t\ne 0\end{array}$

Properties:

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0) ⇒ x(t) δ(t) = x(0)

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$ 

5. $\underset{-\mathrm{\infty }}{\overset{T}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=0$  If T < t0

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

7. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt={\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

Application:

Given expression,

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-a{t}^2}\delta (t+10)dt}$ can be written as

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-a{t}^2}\delta (t-(-10))dt}$

⇒ Impulse is existing at t = -10

But in the given limits of integration, it does not exist, so from the fifth property mentioned, we can get

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-a{t}^2}\delta (t-(-10))dt}$ = 0

⇒ ${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-a{t}^2}\delta (t+10)dt}$ = 0

Given:

Impulse response, h(t) = e^-5t u(t)

output , Y(t) = e^-3t u(t) – e^-5t u(t)

Taking Laplace transform of h(t),

 $H\left(s\right)=\frac{1}{s+5}$

Taking Laplace transform of y(t),

$Y\left(s\right)=\frac{1}{s+3}-\frac{1}{s+5}=\frac{2}{\left(s+3\right)\left(s+5\right)}$

Now,

Y(s) = X (s) H(s)

$X\left(s\right)=\frac{Y\left(s\right)}{H\left(s\right)}=\frac{2}{s+3}$

Taking inverse Laplace,

x(t) = 2e^-3t u(t)

Concept:

Time-shifting property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

$x\left(t-t_0\right)\leftrightarrow X\left(\omega \right)e^{-j\omega t_0}$  ---(1)

Duality property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

$X\left(t\right)\leftrightarrow 2\pi x\left(-\omega \right)$

Calculation:

Fourier Transform of δ(t) = 1

Now by definition of inverse Fourier transform

$\delta (\omega )\leftrightarrow \frac{1}{2\pi }$

Now using time-shifting property from equation (1)

$\delta (\omega -{\omega }_o)\leftrightarrow \frac{1}{2\pi }{e}^{j{\omega }_{o}t}$

Hence option  (3) is the correct answer.

Important Points

Unit Impulse function:

A continuous-time unit impulse function δ(t), also called a Dirac delta function is defined as:

δ (t) = ∞ , t = 0

= 0,  otherwise

The unit impulse function is represented by an arrow with the strength of  ‘1’ which represents its area.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

Properties of Delta function:

Scaling Property:

$\delta \left(at\right)=\frac{1}{\left|at\right|}\delta \left(t\right)$

Multiplication Property:

X(t).δ(t – t0) = x(t0)δ(t – t0)

Sampling Property:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_0\right)dt=x\left(t_0\right)$

Important Expressions:

 X(t).δ(t) = x(0)δ(t)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t\right)dt=x\left(0\right)$

$\mathrm{y}(\mathrm{n})=\sum _{\mathrm{n}=-5}^5\sin (2\mathrm{n})\delta (\mathrm{n}+7)$

n + 7 = 0

n = -7

which does not lie in between -5 ≤ n ≤ 5

hence y[5] = 0

Concept:

Shifting property of convolution

Convolution of the signal x(t) with the impulse gives the same signal as a result after the convolution.

x(t) ∗ δ(t) = x(t)

x(t) ∗ δ(t – t0) = x(t – t0)

Substitute ‘t – t0’ in the place of ‘t’

calculation:

Given:

x(n) = {1, 1, 1, 1, 0.5, 0.5}

y(n) = conv (δ(n – 1), x(n)) 

Using property of impilse function:

y(n) = x(n - 1)

Properties of δ(t):

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0) δ(t – t0)

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$

5. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt={\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

Properties of δ(n):

1. x[n] δ[n] = x[0] δ[n]

2. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n\right]$

$=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[0\right]\delta \left[n\right]=x\left[0\right]$

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n-n_0\right]$

$=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n_0\right]\delta \left[n-n_0\right]=x\left[n_0\right]$

5. δ[an] = δ[n]

Concept:

Unit Impulse function:

A continuous-time unit impulse function δ(t), also called a Dirac delta function is defined as:

δ (t) = ∞ , t = 0

= 0,  otherwise

The unit impulse function is represented by an arrow with the strength of  ‘1’ which represents its area.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

Properties of Delta function:

Scaling Property:

$\delta \left(at\right)=\frac{1}{\left|at\right|}\delta \left(t\right)$

Multiplication Property:

X(t).δ(t – t₀) = x(t₀)δ(t – t₀)

Sampling Property:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_0\right)dt=x\left(t_0\right)$

Important Expressions:

 X(t).δ(t) = x(0)δ(t)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t\right)dt=x\left(0\right)$

Explanation:

From the concept part we can find that:

option(1) is true

From option (2)

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)\delta (\tau )d\tau =1}$

This option is false

As

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta (\tau )d\tau =1}$

Both options (3) and (4) follow delta function properties,

Hence they are true.

Important Points

Differentiation Property:

$\underset{t_1}{\overset{t_2}{\int }}x\left(t\right){\delta }^n\left(t-t^0\right)dt={\left(-1\right)}^{n}x\left(t_0\right)$

Even Signal property:

δ(-t) = δ(t)

Convolution Property:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(\tau \right)\delta \left(t-\tau \right)d\tau =x\left(t\right)$

Integration Property:

$\underset{t_1}{\overset{t_2}{\int }}x\left(t\right)\delta \left(t\right)dt=x\left(0\right)$ for  (t₁ < t₂)

0, otherwise.

Concept:

Unit Impulse function:

A continuous-time unit impulse function δ(t), also called a Dirac delta function is defined as:

$\delta \left(t\right)=\{\begin{array}{c}\mathrm{\infty },t=0\\ 0,t\ne 0\end{array}$

The unit impulse function is represented by an arrow with the strength of  ‘1’ which represents its area.

Area =

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

Properties of δ(t):

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0)δ(t – t0) 

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$

5. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt=(-1{)}^n{\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

Even Function : x(t) = x(-t)

Odd Function: x(t) = - x(-t)

Explanation:

δ (t) is an even function from its definition, so δ(t) = δ(- t)

From property, 

$\delta (2t)=\frac{1}{|2|}\delta (t)=\frac{1}{2}\delta (t)$

Important Points.

Properties of δ(n):

1. x[n] δ[n] = x[0] δ[n]

2. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[0\right]\delta \left[n\right]=x\left[0\right]$

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. $\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n\right]\delta \left[n-n_0\right]=\sum _{n=-\mathrm{\infty }}^{n=\mathrm{\infty }}x\left[n_0\right]\delta \left[n-n_0\right]=x\left[n_0\right]$

5. δ[an] = δ[n]

Concept:

Unit impulse function:

It is defined as, $\delta \left(t\right)=\{\begin{array}{c}\mathrm{\infty },t=0\\ 0,t\ne 0\end{array}$

The discrete-time version of the unit impulse is defined by

$\delta \left[n\right]=\{\begin{array}{c}1,n=0\\ 0,n\ne 0\end{array}$

Properties:

1. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)dt=1$

2. $\delta \left(at\right)=\frac{1}{\left|a\right|}\delta \left(t\right)$

3. x(t) δ(t – t0) = x(t0)

4. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)\delta \left(t-t_o\right)dt=x\left(t_0\right)$

5. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\delta \left(at+b\right)dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(t\right)\frac{1}{\left|a\right|}\delta \left(t+\frac{b}{a}\right)dt$

6. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right){\delta }^n\left(t-t_o\right)dt={\frac{d^{n}x}{d{t}^n}|}_{t=t_0}$

Calculation:

$Lety={\int }_{-7}^2\left(t^2+t^3+1\right)\delta \left(t-3\right)dt$

Using property (3) in the above equation:

As t = 3, lies outside the limit

so y = 0 or we can say that

${\int }_{-7}^2\left(t^2+t^3+1\right)\delta \left(t-3\right)dt=0$

Hence option (4) is the correct answer.