Question 1 |
A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of 1 \mu \mathrm{s}. The time for sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of 1 \mathrm{~V} with 12-bit accuracy is ____ \mu \mathrm{s}. (rounded off to two decimal places)
2.32 | |
6.65 | |
8.32 | |
12.36 |
Question 1 Explanation:
Given: Time constant (\tau) of 1 \mu \mathrm{sec}.
Full scale voltage =1 \mathrm{~V}
The voltage across capacitor is given as
\quad \quad V_{c}(t)=V_{\text {in }}\left(1-e^{-t / \tau}\right)
\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)
To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by
\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right] \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}
Comparing equations (i) and (ii), we get,
\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}
Full scale voltage =1 \mathrm{~V}
The voltage across capacitor is given as
\quad \quad V_{c}(t)=V_{\text {in }}\left(1-e^{-t / \tau}\right)
\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)
To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by
\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right] \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}
Comparing equations (i) and (ii), we get,
\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}
Question 2 |
The S-parameters of a two port network is given as
[S]=\left[\begin{array}{ll} S_{11} & S_{12} \\ S_{21} & S_{22} \end{array}\right]
with reference to Z_{0}. Two lossless transmission line sections of electrical lengths \theta_{1}=\beta l_{1} and \theta_{2}=\beta l 2 are added to the input and output ports for measurement purposes, respectively. The S-parameters \left[S^{\prime}\right] of the resultant two port network is
[S]=\left[\begin{array}{ll} S_{11} & S_{12} \\ S_{21} & S_{22} \end{array}\right]
with reference to Z_{0}. Two lossless transmission line sections of electrical lengths \theta_{1}=\beta l_{1} and \theta_{2}=\beta l 2 are added to the input and output ports for measurement purposes, respectively. The S-parameters \left[S^{\prime}\right] of the resultant two port network is
\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right] | |
\left[\begin{array}{cc}S_{11} e^{j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right] | |
\left[\begin{array}{cc}S_{1} e^{j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right] | |
\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right] |
Question 2 Explanation:
Let us evaluate S_{11} and S_{21} first at V_{2}^{+}=0
(A) S11:
\begin{aligned} V_{1}&=V_{1}^{+}+V_{1}^{-}=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}+\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ \therefore \quad V_{1}^{+}&=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}} \\ V_{1}^{-}&=\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ S_{11}&=\frac{V_{1}^{-}}{V_{1}^{+}}=\frac{\left(V_{1}^{-}\right) e^{+j \beta l_{1}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{11}^{\prime} e^{+j 2 \beta l_{1}} \\ \Rightarrow \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \beta l_{1}}=S_{11} e^{-j 2 \theta_{1}} \\ \therefore \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \theta_{1}} \end{aligned}
(B) S_{21} :
V_{2}=V_{2}^{+} e^{+j \beta l_{2}}+V_{2}^{-} e^{-j \beta l_{2}}=\left(V_{2}^{+}\right)^{\prime}+\left(V_{2}^{-}\right)^{\prime}
Here, V_{2}^{+}=0
Hence,
V_{2}=\left(V_{2}^{-}\right)^{\prime}=V_{2}^{-} e^{-j \beta l_{2}}
\Rightarrow \quad V_{2}^{-}=\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}
From previous discussion in S_{11},
\begin{aligned} V_{1}^{+}&=\left(V_{1}^{+}\right) e^{-j l_{1}} \\ \therefore \quad S_{21}&=\frac{V_{2}^{-}}{V_{1}^{+}}=\frac{\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{21}^{\prime} e^{+j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow S_{21}^{\prime}&=S_{21} e^{-j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow \quad S_{21}^{\prime}&=S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} \end{aligned}
(A) S11:
\begin{aligned} V_{1}&=V_{1}^{+}+V_{1}^{-}=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}+\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ \therefore \quad V_{1}^{+}&=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}} \\ V_{1}^{-}&=\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ S_{11}&=\frac{V_{1}^{-}}{V_{1}^{+}}=\frac{\left(V_{1}^{-}\right) e^{+j \beta l_{1}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{11}^{\prime} e^{+j 2 \beta l_{1}} \\ \Rightarrow \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \beta l_{1}}=S_{11} e^{-j 2 \theta_{1}} \\ \therefore \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \theta_{1}} \end{aligned}
(B) S_{21} :
V_{2}=V_{2}^{+} e^{+j \beta l_{2}}+V_{2}^{-} e^{-j \beta l_{2}}=\left(V_{2}^{+}\right)^{\prime}+\left(V_{2}^{-}\right)^{\prime}
Here, V_{2}^{+}=0
Hence,
V_{2}=\left(V_{2}^{-}\right)^{\prime}=V_{2}^{-} e^{-j \beta l_{2}}
\Rightarrow \quad V_{2}^{-}=\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}
From previous discussion in S_{11},
\begin{aligned} V_{1}^{+}&=\left(V_{1}^{+}\right) e^{-j l_{1}} \\ \therefore \quad S_{21}&=\frac{V_{2}^{-}}{V_{1}^{+}}=\frac{\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{21}^{\prime} e^{+j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow S_{21}^{\prime}&=S_{21} e^{-j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow \quad S_{21}^{\prime}&=S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} \end{aligned}
Question 3 |
The h-parameters of a two port network are shown below. The condition for the maximum small signal voltage gain \frac{V_{\text {out }}}{V_{s}} is
h_{11}=0, h_{12}=0, h_{21}= very high and h_{22}=0 | |
h_{11}= very high, h_{12}=0, h_{21}= very high and h_{22}=0 | |
h_{11}=0, h_{12}= very high, h_{21}= very high and h_{22}=0 | |
h_{11}=0, h_{12}=0, h_{21}= very high and h_{22}= very high |
Question 3 Explanation:
Dependent current source should have h_{21} I_{1} instead of h_{21} V_{1} according to h-parameter.
A_{V}=\frac{V_{\text {out }}}{V_{s}}=\frac{-h_{21} I_{1} \times\left(\frac{1}{h_{22}} \| R_{L}\right)}{h_{11} I_{1}+h_{12} V_{2}}
To achieve maximum \frac{V_{\text {out }}}{V_{s}}
\begin{aligned} & h_{11}=0, \quad h_{12}=0 \\ & h_{21}=\text { Very high, } \quad h_{22}=0 \end{aligned}
Hence, answer should be (A) according to h_{21} I_{1}.
A_{V}=\frac{V_{\text {out }}}{V_{s}}=\frac{-h_{21} I_{1} \times\left(\frac{1}{h_{22}} \| R_{L}\right)}{h_{11} I_{1}+h_{12} V_{2}}
To achieve maximum \frac{V_{\text {out }}}{V_{s}}
\begin{aligned} & h_{11}=0, \quad h_{12}=0 \\ & h_{21}=\text { Very high, } \quad h_{22}=0 \end{aligned}
Hence, answer should be (A) according to h_{21} I_{1}.
Question 4 |
The switch S_{1} was closed and S_{2} was open for a long time. At t=0, switch S_{1} is opened and S_{2} is closed, simultaneously. The value of i_{c}\left(0^{+}\right), in amperes, is
1 | |
-1 | |
0.2 | |
0.8 |
Question 4 Explanation:
At t=0^{-} ; S_{1} \rightarrow closed, S_{2} \rightarrow opened
\begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}
At t=0^{+} ; S_{1} \rightarrow opened, S_{2} \rightarrow closed
\begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}
\begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}
At t=0^{+} ; S_{1} \rightarrow opened, S_{2} \rightarrow closed
\begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}
Question 5 |
In the circuit shown below, switch S was closed for a long time. If the switch is opened at t=0, the maximum magnitude of the voltage V_{R}, in volts, is ____ (rounded off to the nearest integer).
2 | |
4 | |
6 | |
8 |
Question 5 Explanation:
At t=0^{-}
i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}
At t=0^{+}
V_{R}=-2 \times 2=-4,
Magnitude of voltage V_{R},
\left|V_{R}\right|=4
i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}
At t=0^{+}
V_{R}=-2 \times 2=-4,
Magnitude of voltage V_{R},
\left|V_{R}\right|=4
There are 5 questions to complete.