Problem set 23

1. A diode $D$ as shown in the circuit has an $i-V$ relation that can be approximated by $$\begin{align*} i_{_D}=\begin{cases} v^2_{_D}+2v_{_D},&\text{for }v_{_D}>0\\ 0,&\text{for }v_{_D}\leq 0 \end{cases} \end{align*}$$
   
   The value of $v_{_D}$ in the circuit is
1. $(-1+\sqrt{11})$
2. $8V$
3. $5V$
4. $2V$
Applying Kirchhoff's voltage law $$i_{_D}(1)+v_{_D}=10$$ Also, for $v_{_D}>0$ we have $$i_{_D}=v^2_{_D}+2v_{_D}$$ Substituting value of $i_{_D}$ from second equation into the first equation, we get $$v^2_{_D}+3v_{_D}-10=0$$ Solving this quadratic equation for $v_{_D}$, we get, $v_{_D}=2$ or $v_{_D}=-5$. But for $v_{_D}<0$ we get $i_{_D}=0$, hence $v_{_D}=2$.
Hence, answer is (D)
2. The Taylor expansion of the function $\ln{(\cosh x)}$, where $x$ is real, about the point $x= 0$ starts with the following terms:
1. $-\frac{1}{2}x^2+\frac{1}{12}x^4+\cdots$
2. $\frac{1}{2}x^2-\frac{1}{12}x^4+\cdots$
3. $-\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
4. $\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
We will require following identities: $$\sinh{x}=\frac{e^x-e^{-x}}{2}$$ $$\cosh{x}=\frac{e^x+e^{-x}}{2}$$ $$\tanh{x}=\frac{\sinh{x}}{\cosh{x}}$$ $$\frac{d}{dx}\sinh{x}=\cosh{x}$$ $$\frac{d}{dx}\cosh{x}=\sinh{x}$$ $$\frac{d}{dx}\tanh{x}=\text{sech }^2{x}$$ $$\frac{d}{dx}\text{sech }{x}=-\text{sech }{x}\tanh{x}$$ Taylor series of $f(x)$ about $x=0$ is given by $$\begin{align*} f(x)&=f(x)+\frac{1}{1!}f'(0)x+\frac{1}{2!}f''(0)x^2\\ &+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f''''(0)x^4+.. \end{align*}$$ $$\begin{align*} &f(x)=\ln{(\cosh{x})}\\ &\Rightarrow f(0)=0\\ &f'(x)=\tanh{x} \\ &\Rightarrow f'(0)=0\\ &f''(x)=\text{sech }^2{x}=1-\tanh^2{x}\\ &\Rightarrow f''(0)=1\\ &f'''(x)=-2\text{sech }^2{x}\tanh{x}\\ &\Rightarrow f'''(0)=0\\ f''''(x)&=4\text{sech }^2{x}\tanh^2{x}-2\text{sech }^4{x} \\ &\Rightarrow f''''(0)=-2 \end{align*}$$ $$f(x)=\frac{1}{2!}x^2-\frac{1}{12}x^4+\cdots$$ Hence, answer is (B)
3. Given $2\times2$ unitary matrix $U$ satisfying $U^\dagger U=UU^\dagger = 1$ with $\det U = e^{i\phi}$, one can construct a unitary matrix $V \:( V^\dagger V = VV^\dagger= 1)$ with $\det V=1$ from it by
1. multiplying $U$ by $e^{i\phi/2}$
2. multiplying any single element of $U$ by $e^{i\phi}$
3. multiplying any row or column of $U$ by $e^{i\phi/2}$
4. multiplying $U$ by $e^{i\phi}$
The general $2\times 2$ unitary matrix is defined by $$\begin{equation*} U=\begin{pmatrix} a&b\\ -e^{i\phi}b^*&e^{i\phi}a^* \end{pmatrix} \end{equation*}$$ Clearly, as columns or rows of $U$ form an orthonormal set, we have, $|a|^2+|b|^2=1$ and $\det{U}=e^{i\phi}$.
Let us take option (A) i.e. multiplying $U$ by $e^{-i\phi/2}$ $$\begin{equation*} V=e^{-i\phi}U=\begin{pmatrix} e^{-i\phi/2}a&e^{-i\phi/2}b\\ -e^{i\phi/2}b^*&e^{i\phi/2}a^* \end{pmatrix} \end{equation*}$$ Clearly, we get, $\det{V}=1$.
Hence, option (A) is correct
4. The value of the integral $\int_C \frac{z^3dz}{z^2-5z+6}$, where $C$ is a closed contour defined by the equation $2|z|- 5 = 0$, traversed in the anti-clockwise direction, is
1. $-16\pi i$
2. $16\pi i$
3. $8\pi i$
4. $2\pi i$
$$\begin{align*} &\int_C\frac{z^3dz}{z^2-5z+6} \quad C:|z|=\frac{5}{2}=2.5\\ &= \int_C\frac{z^3dz}{(z-2)(z-3)}\\ &= \int_C\frac{\frac{z^3}{(z-3)}}{(z-2)}dz \end{align*}$$ The function has a pole of order 1 at $z=2$ which lies inside $C$. According to residue theorem $$\begin{align*} &\oint\limits_Cf(z)dz\\ &=2\pi i\times&\text{sum of residues at}\\ &&\text{ poles inside }C \end{align*}$$ In general the residue at a pole of order $m$ at $z=z_0$ is $$\begin{eqnarray} \begin{array}{c}R=\frac{1}{(m-1)!}\lim\limits_{z\to z_0}\left\{\frac{d^{m-1}\left((z-z_0)^mf(z)\right)}{dz^{m-1}}\right\}_{z=z_0}\end{array} \end{eqnarray}$$ For simple pole of order 1 $$R=\lim\limits_{z\to z_0}\left\{(z-z_0)f(z_0)\right\}$$ $$R=\lim\limits_{z\to 2}\left\{(z-2)\frac{\frac{z^3}{(z-3)}}{(z-2)}\right\}=-8$$ $$\oint_Cf(z)dz=2\pi i\times(-8)=-16\pi i$$ Hence, answer is (A)
5. The function $f(x)$ obeys the differential equation $\frac{d^2f}{dx^2}-(3 - 2i)f = 0$ and satisfies the conditions $f(0) = 1$ and $f(x)\rightarrow \infty$ as $x\rightarrow 0$. The value of $f(\pi)$ is
1. $e^{2\pi}$
2. $e^{-2\pi}$
3. $-e^{-2\pi}$
4. $-e^{2\pi i}$