Zero State Response MCQ Quiz - Objective Question with Answer for Zero State Response - Download Free PDF

From the given state space representation,

$A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

= $\left[\begin{array}{cc}s-1& 0\\ -1& s-1\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{{\left(s-1\right)}^2}\left[\begin{array}{cc}s-1& 0\\ 1& s-1\end{array}\right]$

= $\left[\begin{array}{cc}\frac{1}{s-1}& 0\\ \frac{1}{{\left(s-1\right)}^2}& \frac{1}{s-1}\end{array}\right]$

$e^{At}=L^{-1}\left[\begin{array}{cc}\frac{1}{s-1}& 0\\ \frac{1}{{\left(s-1\right)}^2}& \frac{1}{s-1}\end{array}\right]$

= $\left[\begin{array}{cc}{e}^t& 0\\ t{e}^t& e^t\end{array}\right]$

X(t) = e^At X(0)

= $\left[\begin{array}{cc}{e}^t& 0\\ t{e}^t& e^t\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]$

= $\left[\begin{array}{c}{e}^t\\ t{e}^t\end{array}\right]$

M, and N circles:

Constant magnitude loci that are M-circles.

Constant phase angle loci that are N-circles.

These are the fundamental components in designing the Nichols chart.

​N-circle:

The phase angle is 

N = Tan [tan-1(Y/X) - Tan-1(Y/(X-1)) ]

All N-circles intersect the real axis between -1 and origin only.

Note:

​M-circle:

The equation for M- circles 

X2 [M-1] + Y2 [M-1] +2XM2 +M2 = 0 -------- (1)

Case -1: 

If M ≠ 1, the equation (1) represents a family of circles.​

Centre = [(-M2) / (M2-1), 0]
​Radius = M / (M2-1)

Case-2:

If M = 1, The equation (1) represents a straight line.

If M = 1, the M loci symmetrical with the Imaginary axis.

​

Initial conditions: x₁(0) = 0, x₂ (0) = 1

ẋ₁(t) = x₂(t)

By applying the Laplace transform, we get

sx₁(s) – x₁(0) = x₂(s)

⇒ s x­₁(s) = x₂(s)

$\Rightarrow x_1\left(s\right)=\frac{x_2\left(s\right)}{s}$

ẋ₂(t) = -6x₁(t) – 5x₂(t)

sx₂(s) – x₂(0) = -6x₁(s) – 5x₂(s)

$s{x}_2\left(s\right)-1=\frac{-6x_2\left(s\right)}{s}-5x_2\left(s\right)$

$\Rightarrow x_2\left(s\right)=\frac{1}{\left(s+\frac{6}{s}+5\right)}$

$\Rightarrow x_2\left(s\right)=\frac{-2}{s+2}+\frac{3}{s+3}$

By applying the inverse Laplace transform,

x₂(t) = -2e^-2t + 3e^-3t

x₂(1) = -2e^-2 + 3e^-3 = -0.12

Alternate Method:

$\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -6& -5\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]$

Since, ẋ₁(t) = x₂(t) & x₁(0) = 0, x₂(0) = 1

ẋ₂(t) = -6x₁(t) – 5x₂(t)

Now, to calculate ZIR

$\varphi \left(s\right)={\left(sI-A\right)}^{-1}=\frac{Adj\left(sI-A\right)}{\left|sI-A\right|}=\frac{\left[\begin{array}{cc}s+5& 1\\ -6& s\end{array}\right]}{s^2+5s+6}=\left[\begin{array}{cc}\frac{s+5}{\left(s+2\right)\left(s+3\right)}& \frac{1}{\left(s+2\right)\left(s+3\right)}\\ \frac{-6}{\left(s+2\right)\left(s+3\right)}& \frac{s}{\left(s+2\right)\left(s+3\right)}\end{array}\right]$

$\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]={\varphi }_1\left(t\right)x\left(0\right)=\left[\begin{array}{cc}{\varphi }_{11}\left(t\right)& {\varphi }_{12}\left(t\right)\\ {\varphi }_{21}\left(t\right)& {\varphi }_{22}\left(t\right)\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]=\left[\begin{array}{c}{\varphi }_{12}\left(t\right)\\ {\varphi }_{22}\left(t\right)\end{array}\right]$  

$\Rightarrow x_2\left(t\right)={\varphi }_2\left(t\right)=L^{-1}\left\{\frac{s}{\left(s+2\right)\left(s+3\right)}\right\}=-2e^{-2t}+3e^{-3t}$

⇒ x₂(1) = -2e^-2 + 3e^-3 = -0.121

x₂(1) = -0.121  

The total response of the system is given by

X(t) = ZIR + ZSR

Where,

ZIR = Zero input response

ZSR = Zero state response

ZIR = e^t x(0)

ZSR = L^-1 [ϕ(s) B U(s)]

Where

Φ(s) = (SI - A)^-1

Calculations:

ZIR = e^at x(0)

X(0) = $\left[\begin{array}{c}0\\ 0\end{array}\right]$

$\Rightarrow ZIR=\left[\begin{array}{c}0\\ 0\end{array}\right]$

ZSR = L^-1 [ϕ (s) B U(s)]

$SIA=\left[\begin{array}{cc}s& 0\\ 0& s+9\end{array}\right]$

$Adj\left(sI-A\right)=\left[\begin{array}{cc}s+9& 0\\ 0& s\end{array}\right]$

$\varphi \left(S\right)={\left(sI-A\right)}^{-1}=\frac{Adj\left[sI-A\right]}{\left|sI-a\right|}$

$=\left[\begin{array}{cc}\frac{1}{s}& 0\\ 0& \frac{1}{s+9}\end{array}\right]$

$ZSR=L^{-1}\left[\left(\begin{array}{cc}\frac{1}{s}& 0\\ 0& \frac{1}{s+9}\end{array}\right)\left(\begin{array}{c}0\\ 45\end{array}\right)\left(\frac{1}{s}\right)\right]$

CI $=L^{-1}\left[\begin{array}{c}0\\ \frac{45}{s\left(s+9\right)}\end{array}\right]=\left[\begin{array}{c}0\\ 5\left(1-e^{-9t}\right)\end{array}\right]$

x1(t) = 0

x2 (t) = 5 (1 – e-9t)

$\underset{t\to \mathrm{\infty }}{lim}\left|\sqrt{x_1^2\left(t\right)+x_2^2\left(t\right)}\right|$

$=\sqrt{0+5^2}$

= 5

$\dot{x}=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]x,x\left(0\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$

$A=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$

The solution x(t) = e^At x(0)

$e^{At}=L^{-1}\left[{\left(sI-A\right)}^{-1}\right]$

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$ 

$=\left[\begin{array}{cc}s+1& 0\\ 0& s+2\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{\left(s+1\right)\left(s+2\right)}\left[\begin{array}{cc}s+2& 0\\ 0& s+1\end{array}\right]$ 

$=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$

$e^{At}=L^{-1}\left[{\left(sI-A\right)}^{-1}\right]=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]$

$x\left(t\right)=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]$

The total response of the system is given by

X(t) = ZIR + ZSR

Where,

ZIR = Zero input response

ZSR = Zero state response

ZIR = e^t x(0)

ZSR = L^-1 [ϕ(s) B U(s)]

Where

Φ(s) = (SI - A)^-1

Calculations:

ZIR = e^at x(0)

X(0) = $\left[\begin{array}{c}0\\ 0\end{array}\right]$

$\Rightarrow ZIR=\left[\begin{array}{c}0\\ 0\end{array}\right]$

ZSR = L^-1 [ϕ (s) B U(s)]

$SIA=\left[\begin{array}{cc}s& 0\\ 0& s+9\end{array}\right]$

$Adj\left(sI-A\right)=\left[\begin{array}{cc}s+9& 0\\ 0& s\end{array}\right]$

$\varphi \left(S\right)={\left(sI-A\right)}^{-1}=\frac{Adj\left[sI-A\right]}{\left|sI-a\right|}$

$=\left[\begin{array}{cc}\frac{1}{s}& 0\\ 0& \frac{1}{s+9}\end{array}\right]$

$ZSR=L^{-1}\left[\left(\begin{array}{cc}\frac{1}{s}& 0\\ 0& \frac{1}{s+9}\end{array}\right)\left(\begin{array}{c}0\\ 45\end{array}\right)\left(\frac{1}{s}\right)\right]$

CI $=L^{-1}\left[\begin{array}{c}0\\ \frac{45}{s\left(s+9\right)}\end{array}\right]=\left[\begin{array}{c}0\\ 5\left(1-e^{-9t}\right)\end{array}\right]$

x1(t) = 0

x2 (t) = 5 (1 – e-9t)

$\underset{t\to \mathrm{\infty }}{lim}\left|\sqrt{x_1^2\left(t\right)+x_2^2\left(t\right)}\right|$

$=\sqrt{0+5^2}$

= 5

$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]=\left[\begin{array}{rr}1& 2\\ -3& -4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}20\\ 10\end{array}\right]$

$A=\left[\begin{array}{cc}1& 2\\ -3& -4\end{array}\right],B=\left[\begin{array}{c}20\\ 10\end{array}\right]$

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{rr}1& 2\\ -3& -4\end{array}\right]$

$=\left[\begin{array}{cc}s-1& -2\\ 3& s+4\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{\left(s-1\right)\left(s+4\right)+6}\left[\begin{array}{cc}s+4& 2\\ -3& s-1\end{array}\right]$

$x\left(t\right)=L^{-1}\left[e^{At}X\left(0\right)+e^{At}.Bu\left(s\right)\right]$

Given that X(0) = 0

$\Rightarrow x\left(t\right)=L^{-1}\left[e^{At}Bu\left(s\right)\right]$

x(s) = e^At B u(s)

$=\frac{1}{s}\left[\frac{1}{\left(s-1\right)\left(s+4\right)+6}\right]\left[\begin{array}{cc}s+4& 2\\ -3& s-1\end{array}\right]\left[\begin{array}{c}20\\ 10\end{array}\right]$

$x\left(\mathrm{\infty }\right)=\underset{s\to 0}{\mathrm{l}\mathrm{t}}sX\left(s\right)$

$=\underset{s\to 0}{\mathrm{l}\mathrm{t}}\frac{1}{\left(s-1\right)\left(s+4\right)+6}\left[\begin{array}{cc}s+4& 2\\ -3& s-1\end{array}\right]\left[\begin{array}{c}20\\ 10\end{array}\right]$

$=\frac{1}{\left(-1\right)\left(4\right)+6}\left[\begin{array}{rr}4& 2\\ -3& -1\end{array}\right]\left[\begin{array}{c}20\\ 10\end{array}\right]$

M, and N circles:

Constant magnitude loci that are M-circles.

Constant phase angle loci that are N-circles.

These are the fundamental components in designing the Nichols chart.

​N-circle:

The phase angle is 

N = Tan [tan-1(Y/X) - Tan-1(Y/(X-1)) ]

All N-circles intersect the real axis between -1 and origin only.

Note:

​M-circle:

The equation for M- circles 

X2 [M-1] + Y2 [M-1] +2XM2 +M2 = 0 -------- (1)

Case -1: 

If M ≠ 1, the equation (1) represents a family of circles.​

Centre = [(-M2) / (M2-1), 0]
​Radius = M / (M2-1)

Case-2:

If M = 1, The equation (1) represents a straight line.

If M = 1, the M loci symmetrical with the Imaginary axis.

​

Initial conditions: x₁(0) = 0, x₂ (0) = 1

ẋ₁(t) = x₂(t)

By applying the Laplace transform, we get

sx₁(s) – x₁(0) = x₂(s)

⇒ s x­₁(s) = x₂(s)

$\Rightarrow x_1\left(s\right)=\frac{x_2\left(s\right)}{s}$

ẋ₂(t) = -6x₁(t) – 5x₂(t)

sx₂(s) – x₂(0) = -6x₁(s) – 5x₂(s)

$s{x}_2\left(s\right)-1=\frac{-6x_2\left(s\right)}{s}-5x_2\left(s\right)$

$\Rightarrow x_2\left(s\right)=\frac{1}{\left(s+\frac{6}{s}+5\right)}$

$\Rightarrow x_2\left(s\right)=\frac{-2}{s+2}+\frac{3}{s+3}$

By applying the inverse Laplace transform,

x₂(t) = -2e^-2t + 3e^-3t

x₂(1) = -2e^-2 + 3e^-3 = -0.12

Alternate Method:

$\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -6& -5\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]$

Since, ẋ₁(t) = x₂(t) & x₁(0) = 0, x₂(0) = 1

ẋ₂(t) = -6x₁(t) – 5x₂(t)

Now, to calculate ZIR

$\varphi \left(s\right)={\left(sI-A\right)}^{-1}=\frac{Adj\left(sI-A\right)}{\left|sI-A\right|}=\frac{\left[\begin{array}{cc}s+5& 1\\ -6& s\end{array}\right]}{s^2+5s+6}=\left[\begin{array}{cc}\frac{s+5}{\left(s+2\right)\left(s+3\right)}& \frac{1}{\left(s+2\right)\left(s+3\right)}\\ \frac{-6}{\left(s+2\right)\left(s+3\right)}& \frac{s}{\left(s+2\right)\left(s+3\right)}\end{array}\right]$

$\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]={\varphi }_1\left(t\right)x\left(0\right)=\left[\begin{array}{cc}{\varphi }_{11}\left(t\right)& {\varphi }_{12}\left(t\right)\\ {\varphi }_{21}\left(t\right)& {\varphi }_{22}\left(t\right)\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]=\left[\begin{array}{c}{\varphi }_{12}\left(t\right)\\ {\varphi }_{22}\left(t\right)\end{array}\right]$  

$\Rightarrow x_2\left(t\right)={\varphi }_2\left(t\right)=L^{-1}\left\{\frac{s}{\left(s+2\right)\left(s+3\right)}\right\}=-2e^{-2t}+3e^{-3t}$

⇒ x₂(1) = -2e^-2 + 3e^-3 = -0.121

x₂(1) = -0.121  

$\dot{x}=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]x,x\left(0\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$

$A=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$

The solution x(t) = e^At x(0)

$e^{At}=L^{-1}\left[{\left(sI-A\right)}^{-1}\right]$

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$ 

$=\left[\begin{array}{cc}s+1& 0\\ 0& s+2\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{\left(s+1\right)\left(s+2\right)}\left[\begin{array}{cc}s+2& 0\\ 0& s+1\end{array}\right]$ 

$=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$

$e^{At}=L^{-1}\left[{\left(sI-A\right)}^{-1}\right]=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]$

$x\left(t\right)=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]$

Taking z-transform of the differential equation:

3[Y(z) z^-2 + y(-1)z^-1 + y(-2)] - 4[Y(z) z^-1 + y(-1)] + Y(z) = X(z)

∴ Y(z)(3z^-2 – 4z^-1 + 1) = X(z) - 6

$Y\left(z\right)=\frac{X\left(z\right)}{3z^{-2}-4z^{-1}+1}-\frac{6}{3z^{-2}-4z^{-1}+1}$

Where the first term on the right hand side represents the Zero State Response and second term represents Zero Input Response.

With x(n) = u(n), the z-transform will be:

$X\left(z\right)=\frac{z}{z-1}$

$\therefore Y\left(z\right)=\frac{z}{\left(z-1\right)\left(3z^{-2}-4z^{-1}+1\right)}-\frac{6}{3z^{-2}-4z^{-1}+1}$

Zero state response = $\frac{z}{\left(z-1\right)\left(3z^{-2}-4z^{-1}+1\right)}$

Zero input response = $-\frac{6}{3z^{-2}-4z^{-1}+1}$

Now, the ratio of the zero state response to the zero input response $R(z)$ will be:

$R\left(z\right)=\frac{\frac{z}{(z-1)\left(3z^{-2}-4z^{-1}+1\right)}}{-\frac{6}{3z^{-2}-4z^{-1}+1}}=\frac{-z}{6\left(z-1\right)}$

$R\left(z\right)=\frac{-1}{6\left(1-z^{-1}\right)}\stackrel{Z{T}^{-1}}{\to }\frac{-1}{6}u\left[n\right]$

$A=\left[\begin{array}{cc}-2& 0\\ 0& -3\end{array}\right]$

x(t) = e^At X(0)

e^At = L^-1 [sI - A]^-1

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-2& 0\\ 0& -3\end{array}\right]$

$=\left[\begin{array}{cc}s+2& 0\\ 0& s+3\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{\left(s+2\right)\left(s+3\right)}\left[\begin{array}{cc}\left(s+3\right)& 0\\ 0& s+2\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{s+2}& 0\\ 0& \frac{1}{s+3}\end{array}\right]$

$L^{-1}\left[{\left(sI-A\right)}^{-1}\right]=\left[\begin{array}{cc}{e}^{-2t}& 0\\ 0& e^{-3t}\end{array}\right]$

$x\left(t\right)=\left[\begin{array}{cc}{e}^{-2t}& 0\\ 0& e^{-3t}\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}{e}^{-2t}\\ e^{-3t}\end{array}\right]$

$y\left(t\right)=Cx\left(t\right)=\left[11\right]\left[\begin{array}{c}{e}^{-2t}\\ e^{-3t}\end{array}\right]=e^{-2t}+e^{-3t}$