Problem set 87

1. Eigenvalues of the matrix $\begin{bmatrix}1&-1\\1&1\end{bmatrix}$
1. 1, -1
2. -1, $-i$
3. $i$, $-i$
4. $1+i$, $1-i$

Determinant of matrix is equal to product of eigenvalues. $$\begin{vmatrix}1&-1\\1&1\end{vmatrix}=2$$ Clearly $$(1+i)(1-i)=2$$

Hence, answer is (D)

2. Consider an $n$-MOSFET with the following parameters: current drive strength $K= 60 \:\mu A/V^2$, breakdown voltage $BV_{DS}=10\: V$, ratio of effective gate width to the channel length $\frac{W}{L}=5$ and threshold voltage $V_{th}=0.5V$. In the circuit given below, this $n$-MOSFET is operating in the
   
1. ohmic region
2. cut-off region
3. saturation region
4. breakdown region

The MOSFET is in

saturation when $V_{GS} > V_{th}$ and $V_{DS} > V_{GS} - V_{th}$.

cut-off region when $V_{GS} < V_{th}$

linear when $V_{GS} > V_{th}$ and $V_{DS} < V_{GS} - V_{th}$.

$V_{GS}=1.2 V$, $V_{th}=0.5V$, $V_{DS}=10 V$

Hence, answer is (C)

3. Particular integral of first order linear differential $\frac{dy}{dx}=x+y$ is given by:
   1. $y(x)=-x-1$
   2. $y(x)=x+1$
   3. $y(x)=x-1$
   4. $y(x)=-x+1$
   Linear differential equation is given by $$\frac{dy}{dx}+P(x)y=Q(x)$$ Multiply both sides by the Integrating Factor $e^{\int P(x)dx}$ $${\scriptstyle \frac{dy}{dx}e^{\int P(x)dx}+P(x)ye^{\int P(x)dx}=Q(x)e^{\int P(x)dx}}$$ $$\frac{d\left(ye^{\int P(x)dx}\right)}{dx}=Q(x)e^{\int P(x)dx}$$ On integrating we get $$ye^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx+c$$ $$\frac{dy}{dx}-y=x$$ $P(x)=-1$, $Q(x)=x$ $$ye^{-\int dx}=\int xe^{-\int dx}dx+c$$ $$ye^{-x}=\int xe^{-x}dx+c$$ Integrating by parts $$ye^{-x}=-xe^{-x}+\int e^{-x}dx+c$$ $$ye^{-x}=-xe^{-x}-e^{-x}+c$$ $$y=-x-1$$

   Hence, answer is (A)

4. The Fourier transform of a Gaussian function is of the form:
1. Exponential
2. Lorentzian
3. Gaussian
4. Screened coulomb

Hence, answer is (C)

5. The real part of $\log{(3+4i)}$ is :
1. $\log2$
2. $\log3$
3. $\log4$
4. $\log5$

Let $z=x+iy$, $r=\sqrt{x^2+y^2}$, $\theta=\tan^{-1}{\frac{y}{x}}$ and $z=re^{i\theta}$

For $z=3+4i$, $r=\sqrt{3^2+16^2}=5$, $\theta=\tan^{-1}{\frac{4}{3}}=0.92729$ and $z=5e^{i0.92729}$ $$\begin{align*} \log{(3+4i)}&=\log{5e^{i0.92729}}\\ &=\log5+i0.92729 \end{align*}$$

Hence, answer is (D)