CnN CLS Notes S2 EndsM PDF

## 7/3/24 ### Initial & Final Value Theorem: For $f(t)$ function of time. * **IVT**: $f(0+) = lim_{s\to 0} sF(s)$ * **FVT**: $f(t) = lim_{s\to\infty} sF(s)$ ### Q) Without finding $f(t)$ first determine $f(0+)$ & $f(\infty)$ for the following transforms: **(a)** $\frac{(2s+1)+4}{[(s+3)+16][(s+...

## 7/3/24 ### Initial & Final Value Theorem: For $f(t)$ function of time. * **IVT**: $f(0+) = lim_{s\to 0} sF(s)$ * **FVT**: $f(t) = lim_{s\to\infty} sF(s)$ ### Q) Without finding $f(t)$ first determine $f(0+)$ & $f(\infty)$ for the following transforms: **(a)** $\frac{(2s+1)+4}{[(s+3)+16][(s+5)]} = \frac{4s^2+4s+5}{s(s^2+6s+25)(s+5)}$ **Using IVT**: $f(0+) = lim_{s\to 0} sF(s) = lim_{s\to 0} \frac{(4s^2+4s+5)s}{s(s^2+6s+25)(s+5)}$ $= lim_{s\to 0} \frac{(4+4/s + 5/s^2)}{(1 + 6/s + 25/s^2)(1 + 5/s)} = 4$ **Using FVT**: $f(t\to\infty) = lim_{s\to \infty} sF(s) = lim_{s\to \infty} \frac{(4s^2 + 4s + 5)s}{s(s^2+6s+25)(s+5)} = 0$ ## Q) Find $i(t)$. Apply IVF & FVT to $i(s)$. **IVF**: $i(0+) = lim_{s\to 0} sI(s) = lim_{s\to 0} \frac{50(10s^2+3s)}{125s^2+45s+2} = \frac{50 \times 10^4}{125 \times 10^4} = 4$ **FVT**: $i(t\to\infty) = lim_{s\to \infty} sI(s) = lim_{s\to \infty} \frac{50(10s^2+3s)}{125s^2+45s+2} = 0$ ## 11/3/24 ### Transfer Functions $H(s) = \frac{Y(s)}{X(s)}$ Where $X(s)$ → Excitation $Y(s)$ → Response $Y(s) = L{y(t)}$ $X(s) = L{x(t)}$ ### Eg:  If the excitation is $v_i(t)$ and response is $v_o(t)$ $H(s) = \frac{V_o(s)}{V_i(s)}$. If the excitation is $v_i(t)$ and response is $i_o(t)$: $H(s) = \frac{I_o(s)}{V_i(s)}$. *(i)* **Excitation:** $v_i(t)$ * **Response:** $v_o(t)$ → Voltage gain $H(s) = \frac{V_o(s)}{V_i(s)}$ *(ii)* **Excitation:** $i_i(t)$ * **Response:** $i_o(t)$ → Current gain $H(s) = \frac{I_o(s)}{I_i(s)}$ *(iii)* Excitation = $i_i(t)$ Response = $i_o(t)$ → Admittance $H(s) = \frac{I_o(s)}{V_i(s)}$ *(iv)* Excitation = $i_i(t)$ Response = $v_o(t)$ → Impedance $H(s) = \frac{V_o(s)}{I_i(s)}$ *When $i(t) = \delta(t)$, then $y(t) = h(t)$ is known as unit impulse response & denoted by $h(t)$.* * $H(s) = \frac{Y(s)}{X(s)} = \frac{Y(s)}{1} = Y(s) //$ ### Qn: The output of a linear system $y(t) = 10 cos(w_o t)$ with the input of $e^{w_o t}u(t)$. Find the transfer function of the system & its impulse response. $X(s) = \frac{1}{s+1}$ $Y(s) = \frac{10 (s+1)}{(s+1)^2 + 4^2}$ $H(s) = \frac{Y(s)}{X(s)} = \frac{10 (s+1)}{(s+1)^2 + 4^2} \times \frac{s+1}{1} = \frac{10 (s+1)}{(s+1)^2 + 4^2}$ $h(t) = L^{-1} {H(s)} = 10 e^{-t} cos(4t)u(t)$ ## 12/8/24 ### Q) Find (a) the TF $H(s) = \frac{V_o}{V_i}$, (b) the amplitude of the impulse response, (c) The response when $V_i(t) = u(t)$, (d) The response when $V_i(t) = 8 cos(2t) V$. **(a)** $V_{ab} = V_i(Z_{ab})$ $Z_{ab}+1$ $Z_{ab} = \frac{(s+1)}{s+2}$ $V_o = V_{ab}(1) = \frac{(s+1)}{s+2} \times V_i\frac{(s+1)}{(s+2)(s+1)+1} = \frac{V_i(s+1)(s+1)}{(s+2)(s+1)+1}$ $V_o = \frac{V_i(s+1)^2}{(s+2)(s+1)+1} = \frac{V_i(s^2+3s+1)}{s^2+3s+2}$ $\frac{V_o}{V_i} = H(s) = \frac{s^2+3s+1}{s^2+3s+2}$ **(b)** Impulse Response: $H(s) = \frac{2}{s(s+2)} \Rightarrow \frac{2}{s(s+2)} h(t)$ $h(t) = 2 \times e^{-2t/2} u(t) = e^{-2t}u(t)$ $A(t) = \frac{1}{2} \Rightarrow A = \frac{1}{2}$ **(c)** $V_i (t) = u(t)$ $V_i(s) = \frac{1}{s}$ $V_o(s) = \frac{V_i(s)}{s(s+2)}$ $V_o(t) = \frac{1}{2} - \frac{1}{2}e^{-2t} u(t)$ **(d)** $V_i(t) = 8 cos(2t)$ $V_i(s) = \frac{8s}{s^2+4}$ $V_o(s) = \frac{8s}{s^2+4} \times \frac{s^2+3s+1}{s^2+3s+2}$ $V_o(t) = (-24 e^{-t}+ \frac{24}{25} cos(2t) + \frac{32}{25} sin(2t))u(t)$ ## 14/3/24 ### Q) $V_i(t) = 8 cos(2t) V$ $V_i(s) = \frac{8s}{s^2+4}$ $V_o(s) = \frac{8s}{s^2+4} \times \frac{2}{s^2+5} = \frac{16s}{(s^2+4)(s^2+5)}$ $V_o(s) = \frac{A}{s+2j} + \frac{B}{s-2j} + \frac{C}{s+j\sqrt{5}} + \frac{D}{s-j\sqrt{5}}$ $A(s+2j)(s+j\sqrt{5})(s-j\sqrt{5}) + B(s-2j)(s+j\sqrt{5})(s-j\sqrt{5}) + C(s+2j)(s-2j)(s-j\sqrt{5}) + D(s+2j)(s-2j)(s+j\sqrt{5}) = 16s$ $A(s^3 + 2js^2 + j\sqrt 5s^2 + 2\sqrt 5s - 5) + B(s^3 - 2js^2 + j\sqrt 5s^2 -2\sqrt 5s - 5) + C(s^3 - 5s - 2js + 10j) + D(s^3 - 5s + 2js - 10j) = 16s$ Equating coefficients and solving for $A, B, C, D$, we get $A = \frac{8}{5}, B = -\frac{8}{5}, C = \frac{4}{5}, D=\frac{4}{5}$ $V_o(t) = (\frac{8}{5} e^{-2t} + \frac{8}{5} e^{2t} + \frac{4}{5} e^{-t\sqrt{5}} + \frac{4}{5} e^{t\sqrt{5}})u(t)$ ## Network Stability $h(t)$ → unit impulse response $H(s) = \frac{N(s)}{D(s)}$ For a network to be stable $lim_{t\to\infty}|h(t)|$ = finite. When degree of $N(s)$ is greater than $D(s)$: $H(s) = K_o s^n + ... K_1 s + K_o + R(s)$ (By Long Division) $H(s) = N(s) / D(s)$ * Due to these, $lim_{t\to\infty} |h(t)|$ → ∞ * The network won't be stable. **Conditions:** 1. **$deg(N(s)) < deg(D(s))$** : The inverse of $H(s)$ does not meet the condition $lim_{t\to\infty} |h(t)|$ = finite. 2. **All poles of $H(s)$ must have negative real parts.** That is all poles must lie on the left half of the s-plane. ### Problems: An active filter has the transform function: $H(s) = \frac{K}{S^2 + S(4 - K) + 1}$ For what values of $K$ is the filter stable? $D(s) = s^2 + s(4-K) + 1$ **Discriminant** $D(s) = b^2 - 4ac = (-4 + K)^2 - 4(1)(1) = 0$ Roots: $\frac{-b \pm \sqrt{b^2-4ac}}{2} = (4-K) \pm \sqrt{(4-K)^2-4}$ = $\frac{-4+K \pm \sqrt{16 - 28K + K^2 - 4}}{2} = \frac{-4+K \pm \sqrt{12 -28K + K^2}}{2}$ If the $\sqrt{12 - 28K + K^2}$ is imaginary, then we can check the condition for real parts. $\frac{-4+K}{2}$ < 0 $4-K < 0$ $K > 4$ If $\sqrt{12 - 28K + K^2}$ is real, $\frac{-4+K \pm \sqrt{12 - 28K + K^2}}{2}$ < 0. $\frac{-b \pm \sqrt{b^2-4ac}}{2}$ < 0 $K < 4$ The above condition $O$ is always true for $b > 0$. $\therefore$ $K < 4$ ## 19/3/24 ### Q) $s^2 + 5(2s + 2) + 25$ $P_{1,2} = \frac{-(4-K)\pm\sqrt{(4-K)^2-4}}{2}$ Considering the underroot as imaginary, $-(2s+2)^2 \le0$ $2s \le 20$ $s \le -25$ Considering the underRoot as Real, $-(2s+2) + \sqrt{(2s+2)^2-100} \le 0$ $-b + \sqrt{b^2-100} \le 0.$ $\therefore$ $s \le -25$ ### Two Port Networks:  A general network having two pairs of terminals is an important building block in electronic systems, transmissions, and distribution systems in which an electrical signal or electrical energy enters the input terminals, is acted upon by the network and leaves through the output terminals. A pair of terminals at which a signal may enter or leave a network is called a port. The currents in the two leads comprising each port must be equal, sources and loads must be connected directly across the two terminals of a port if the methods of analyzing the two port networks are to be used. $i_a = i_b$ $i_c = i_d$  ### Z Parameters: $V_1 = Z_{11}I_1 + Z_{12}I_2$ $V_2 = Z_{21}I_1 + Z_{22}I_2$ With $I_2 = 0$: $V_1 = Z_{11} I_{1}$ → Driving point impedance $Z_{11} = \frac{V_1}{I_1}$ → at $I_1$ with $I_2$ open circuited With $I_1 = 0$ → Open circuit input impedance. **Eg**:  $Z_{11} = \frac{V_1}{I_1}$ with $I_2=0$ Where $V_1 = I_1(2a+2b)$ $V_1 = (2a+2b)I_1 = Z_{11}$ $Z_{21} = \frac{V_2}{I_1}$ with $I_2 = 0$: $Z_{21} = \frac{V_2}{I_1}$ → Transfer Impedance at $I_1$ with $I_2$ open circuited. **Eg:** $Z_{21} = \frac{V_2}{I_1} = Z_b$ with $I_2 = 0$ $V_2 = I_1(Z_b)$ $Z_{22} = \frac{V_2}{I_2}$ with $I_1=0$: $Z_{22} = \frac{V_2}{I_2}$ → Transfer Impedance at $I_2$ with $I_1$ open circuited. **Eg:** $Z_{22} = \frac{V_2}{I_2} = Z_c + Z_b$ $V_2 = I_2(Z_c + Z_b)$ ## 21/3/24 ### Y-Parameters:  $I_1 = Y_{11} V_1 + Y_{12} V_2$ $I_2 = Y_{21} V_1 + Y_{22} V_2$ With $V2= 0$: $I_1 = Y_{11}V_1$ → Input admittance $Y_{11} = \frac{I_1}{V_1}$ with $V_2=0$ With $V_1 = 0$: $I_2 = Y_{21}V_1$ → Transfer Admittance $Y_{21} = \frac{I_2}{V_1}$ with $V_2=0$ With $V_1 = 0$: $I_2 = Y_{22}V_2$ → Output Admittance $Y_{22} = \frac{I_2}{V_2}$ with $V_1=0$ ### Problem:  Find the Y parameters **Step 1:** Short the output $V_2 = 0$ and find $Y_{11}$ and $Y_{21}$  $V_1 = (10/15) \times 51 = 50I_1$ $Y_{11} = \frac{I_1}{V_1} = \frac{15}{50} = 0.3$ $Y_{21} = \frac{I_2}{V_1}$ with $V_2 = 0$ $-I_2 = I(5)$ $I_2 = -\frac{1}{5}V_1$ $Y_{21} = \frac{I_2}{V_1} = -\frac{1}{5} = - 0.1$ **Step 2:** Short the output $V_1 = 0$ and find $Y_{12}$ and $Y_{22}$.  $Y_{12} = \frac{I_1}{V_2}$ with $V_1 = 0$ & $V_2 = -10I_1$ $Y_{12} = \frac{I_1}{-10I_1} = -0.1$ $Y_{22} = \frac{I_2}{V_2}$ with $V_1=0$ and $V_2 = 10I_2$ $Y_{22} = \frac{I_2}{10I_2} = 0.15$ $I_2 = 0.3 V_1 - 0.1 V_2$ $I_2 = - 0.1 V_1 + 0.15 V_2$ ## Connecting Source & Load to the two port Network:  $I_1 = 15 - V_1 = 15 - 0.1V_1$ $I_2 = -V_2 = -0.25V_2$ $I_1 = 0.3V_1 - 0.1V_2 = 15 - 0.1V_1$ $15 = 0.4V_1 - 0.1V_2$ $I_2 = -0.1V_1 + 0.15V_2 = -0.25V_2$ $\Rightarrow 0.1V_1 = 0.4V_2$ $4V_1 - V_2 = 150$ $16V_2-V_2 = 150$ $V_2 = \frac{150}{15}= 10V$ $V_1 = 40V$ $I_1 = 11A, I_2 = -2.5A$ ## 26/3/24 ### Transmission Parameters (ABCD Parameters): $V_1 = AI_2 - BI_1$ $I_1 = CV_2 - DI_2$ $\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_2 \\ I_2 \end{bmatrix}$ $A =\frac{V_1}{I_2}$ with $I_1=0$, $\qquad\ B=\frac{V_1}{I_1}$ with $V_2=0$, $\qquad\ C = \frac{I_1}{V_2}$ with $I_2=0$, $\qquad\ D=\frac{I_1}{I_2}$ with $V_2=0$. ### Problem:  Find the ABCD parameters: $A = \frac{V_1}{I_2}$ with $I_1=0$ $V_2 = 5V_1$ $V_1 = \frac{1}{5}V_2$ $\qquad$ $B= \frac{V_1}{I_1}$ with $V_2=0$. $\qquad$ $V_2 = 5I_1$ (from the above) $\qquad$ $V_1= 17 \Rightarrow B = \frac{17}{5}$ $\qquad$ $C= \frac{I_1}{V_2}$ with $I_2=0$ $\qquad$ $I_1 = \frac{1}{5}I_2$ $\qquad$ $V_2 = 5 \Rightarrow C = \frac{1}{5}$ $\qquad$ $D = \frac{I_1}{I_2}$ with $V_2 = 0$ $\qquad$ $I_1 = \frac{1}{5}I_2$ $\qquad$ $I_2 = \frac{5}{5} = 1 \Rightarrow D = \frac{1}{5}$ ## Consider 2-port Networks connected in cascade  Terminal voltages and currents are indicated for each two port, and the corresponding T parameter relationships: For A: $\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} t_a & t_b \\ t_c & t_d \end{bmatrix} \begin{bmatrix} V_3 \\ I_3 \end{bmatrix}$ For B: $\begin{bmatrix} V_3 \\ I_3 \end{bmatrix} = \begin{bmatrix} t_e & t_f \\ t_g & t_h \end{bmatrix} \begin{bmatrix} V_4 \\ I_4 \end{bmatrix}$ Combining I and II, we get $\begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} t_a & t_b \\ t_c & t_d \end{bmatrix} \begin{bmatrix} t_e & t_f \\ t_g & t_h \end{bmatrix} \begin{bmatrix} V_4 \\ I_4 \end{bmatrix}$ The T parameter matrix is the ABCD parameter for combination of networks. ## 14/3/24 ### N/W A:  $V_2 = V_1(10)$ $I_1 = 0.1V_1$ $A = \frac{V_2}{V_1} = 10$ $C = \frac{I_1}{V_2} = 0.1$ $B = \frac{V_1}{I_1}$ with $V_2=0$ $I_1 = 0.1V_1$ = 0 $V_1 = 0$ * **$B = \frac{0}{0} = \infty$** $D = \frac{I_1}{I_2}$ with $V_2 = 0$ $D = \frac{0.1V_1}{0} = \infty$ ### N/W B:  * $A = \frac{V_2}{V_1} = 10$ * $C = \frac{I_1}{V_2}= 0.1$ * $B = \frac{V_1}{I_1}$ with $V_2=0$ * $V_1 = 6V_2 = 0$ * $B = \frac{0}{0}= \infty$ * $D = \frac{I_1}{I_2}$ with $V_2=0$ * $D = \frac{0.1V_1}{0} = \infty$ ## H.W ### Q)  Find **Z Parameters** * $Z_{11} = \frac{V_1}{I_1}$ with $I_2 = 0$ * $V_1 = Z_{11}I_1 +Z_{12} I_2$ $V_1 = (2+5)I_1 + 5I_2$ **$Z_{11} = 2 + 5 = 7$** **$Z_{12} = 5$** * $Z_{21} = \frac{V_2}{I_1}$ with $I_2 = 0$ $V_2 = Z_{21}I_1 +Z_{22} I_2$ $V_2 = 5 I_1 +(5+3)I_2$ **$Z_{21} = 5$** * $Z_{22} = \frac{V_2}{I_2}$ with $I_1 = 0$ $V_2 = Z_{21}I_1+Z_{22} I_2$ $V_2 = S I_1 +(5+3)I_2$ **$Z_{22} = 5+3 = 8$** ## Q)  Find Z parameters of the second network * $Z_{11} = \frac{V_1}{I_1}$ with $I_2 =0$ $V_1 = V_2 = I_2(4) = 0$ **$Z_{11} = 0$** * $Z_{12} = \frac{V_1}{I_2}$ with $I_1 = 0$ $V_1 = V_2 = I_2(4) = 4I_2$ **$Z_{12} = 4$** * $Z_{21} = \frac{V_2}{I_1}$ with $I_2 = 0$ $V_2 = I_1(2) = 2I_1$ **$Z_{21} = 2$** * $Z_{22} = \frac{V_2}{I_2}$ with $I_1 = 0$ $V_2 = \frac{2I_1}{I_2} = 0$ **$Z_{22}=0$** **$Z = \begin{bmatrix} Z_A & Z_B \\Z_C & Z_D \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 5 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 4 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 7 & 9 \\ 7 & 8 \end{bmatrix}$** ## Verify with actual circuit:  $V_1 = 18I_1 + 7I_2$ from KVL $V_2 = 7I_1 + 28I_2$ $\begin{bmatrix} 18 & 7 \\ 7 & 28 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}$ ## 20/4/20 **Terminated two port network:** Driving port impedance at the input port of the load-terminated network. **Output Impedance:** The output impedance of a network is simply the Thevenin equivalent impedance b/w the output terminals. Assume a specific driving configuration. We shall select an independent voltage source $V_s$ in series with a generator impedance. $V_s = V_1 + I_2 Z_g$ $V_1 = Z_{11}I_1 + Z_{12}I_2$ $V_2 = Z_{21}I_1 + Z_{22} I_2$ $V_1 = V_s - I_2 Z_g$ $V_s - I_2Z_g = Z_{11}I_1 + Z_{12}I_2$ $I_1(Z_{11} + Z_g) = V_s - Z_{12} I_2$ $I_1 = \frac{V_s - Z_{12} I_2}{Z_{11} + Z_g}$ $V_2 = Z_{21} \frac{V_s - Z_{12} I_2}{Z_{11} + Z_g} + Z_{22} I_2$ **Output Impedance**: $Z_{out} = Z{22} - \frac{Z_{12}Z_{21}}{Z_{11} + Z_g}$ If $Z_g = 0$ $Z{out} = Z{22} - \frac{Z_{12}Z_{21}}{Z_{11}}$ **Q)** $\frac{Z_{11} = 10^3 \Omega, Z_{12} = 10^1 \Omega, Z_{21} = 10^1 \Omega, Z_{22} = 10^4 \Omega$. Let the two port network be driven by an ideal sinusoidal voltage source $V_S$ in series with a $500 \Omega$ resistor and terminated in $10 k \Omega$ load. Find voltage gain, current gain $Z_{in}$, $Z_{out}$ **Ans:** 1. $V_1 = Z_{11}I_1 + Z_{12}I_2 = 10^3 I_1 + 10^1I_2$ 2. $V_2 = Z_{21}I_1 + Z_{22}I_2 = 10^1I_1 + 10^4I_2$  $V_s = 500I + V_1$ $V_2 = -10^4 I_2$ $I_1(1500) + 10I_2 = V_s$ $-104 I_2 = -106I + 10^4I_2$ $I_1(-106) + I_2(2\times 10^4) = 0$ $I_1 = \frac{V_s - 10^1I_2}{1500}$ $I_2 (2 \times 10^4) = \frac{10^5V_s - 10^2 I_2}{1500}$ $I_2 (3\times10^7) = 10^5V_s - 10^2I_2$ $I_2(300000) = V_s$ $\therefore I_2 = \frac{V_s}{3 \times 10^5}$ $I_1 = \frac{V_s - 10 (\frac{V_s}{3\times 10^5})}{1500} = \frac{V_s}{1500} - \frac{1}{45\times 10^3} V_s = \frac{1}{1500} - \frac{1}{45 \times 10^3}V_s$ $V_2=-10^4\times \frac{V_S}{3\times 10^5} = -\frac{1}{3}V_s$ $V_o = \frac{-1}{3} V_s$ $V_o = \frac{-1}{3} \times 500 = \frac{-500}{3}V$ Voltage Gain: $A_v = \frac{V_o}{V_i} = \frac{1000}{3} = 333 \frac{1}{3}$ Current Gain: $A_i = \frac{I_2}{I_1} = \frac{\frac{V_s}{3\times10^5}}{\frac{1}{1500} - \frac{1}{45\times 10^3} V_s} = \frac{1}{200}$ $Z_{in} = \frac{V_1}{I_1} = \frac{10^3I_1 + 10^1I_2}{\frac{1}{1500} - \frac{1}{45\times 10^3}V_s} = 1.47 \times 10^3 \Omega$ $Z_{out} = Z_{22} - \frac{Z_{12}Z_{21}}{Z_{11} + Z_g} = 10^4 - \frac{10^1 \times 10^1}{10^3 + 500} = 10^3 \Omega $ ## 5/4/24 $T$ and $\Pi$ representations: A two-port network with any number of elements may be converted into a two port, three-element network. That a two port network may be represented by an equivalent T network i.e.  It is possible to express the elements of the T network in terms of Z parameters or ABCD parameters. ### Z Parameters: $\qquad$ $Z_{11}=\frac{V_1}{I_1}$ with $I_2=0 = Z_a + Z_c$ $\qquad$ $Z_{21}=\frac{V_2