Problem set 84

1. In the schematic figure given below, assume that the propagation delay of each logic gate is $t_{gate}$.
   
   The propagation delay of the circuit will be maximum when the logic inputs A and B make the transition
1. $(0,1)\rightarrow(1,1)$
2. $(1,1)\rightarrow(0,1)$
3. $(0,0)\rightarrow(1,1)$
4. $(0,0)\rightarrow(0,1)$

When input of gate changes its state, the output of gate does not change instantaneously. Instead, the out put changes after a small delay. The delay in the circuit due to all gates is called propagation delay. The states of each gate for different inputs are as shown below.



During transition $(0,1)\rightarrow(1,1)$ from figures (b) and (c) we see that only first OR gate, AND gate, and second OR change their states. Hence, time delay is $3t_{gate}$

During transition $(1,1)\rightarrow(0,1)$ from figures (c) and (b) we see that only first OR gate, AND gate, and second OR change their states. Hence, time delay is $3t_{gate}$

During transition $(0,0)\rightarrow(1,1)$ from figures (a) and (c) we see that only NOT gate changes its state. Hence, time delay is $t_{gate}$

During transition $(0,0)\rightarrow(0,1)$ from figures (a) and (b) we see that all gates change their states. Hence, time delay is $4t_{gate}$

Hence, answer is (D)

2. Given the input voltage $V_i$, which of the following waveforms correctly represents the output voltage $V_0$ in the circuit shown below?
   
1. 
2. 
3. 
4. 

Output voltage is given by $$V_0=A(V_2-V_1)$$ where, $V_1$ and $V_2$ are voltages at inverting and non-inverting inputs and $A$ is gain given by $$A=\frac{R_f}{R_i}=\frac{10}{5}=2$$ $$V_0=2(V_2-V_1)$$ $$V_0=A(0.5-V_1)$$ If $V_1>V_2$, $V_0$ will be negative, but no option have negative voltage.

Hence, $V_1\leq V_2$

Since, $V_2=0.5\: V$, maximum value of $V_1=0.5$, hence, minimum output voltage is $V_{0_{min}}=0$

Also, this is inverting amplifier.

Hence, answer is (B)

3. In finding the roots of the polynomial $f(x)=3x^3-4x-5$ using the iterative Newton-Raphson method, the initial guess is taken to be $x=2$. In the next iteration its value is nearest to
1. 1.671
2. 1.656
3. 1.559
4. 1.551

Newton-Raphson formula is $$\begin{align*} x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=2-\frac{3\times2^3-4\times2-5}{9\times2^2-4}\\ &=1.65625 \end{align*}$$

Hence, answer is (B)

4. For a particle of energy $E$ and $P$ momentum (in a frame $F$), the rapidity $y$ is defined as $y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$. In a frame $F'$ moving with velocity $v=(0,0,\beta c)$ with respect to $F$, the rapidity $y'$ will be
1. $y'=y+\frac{1}{2}\ln{\left(1-\beta^2\right)}$
2. $y'=y-\frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
3. $y'=y+\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
4. $y'=y+2\ln{\left(\frac{1+\beta}{1-\beta}\right)}$

$$y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{E'+p_3'c}{E'-p_3'c}\right)}$$ According to Lorentz transformation (boost) $$p_3'=\gamma\left(p_3-\frac{vE}{c^2}\right)$$ $$E'=\gamma\left(E-vp_3\right)$$ Using $v=\beta c$ $$p_3'=\gamma\left(p_3-\frac{\beta E}{c}\right)$$ $$E'=\gamma\left(E-\beta cp_3\right)$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\gamma\left(E-\beta cp_3\right)+\gamma\left(p_3-\frac{\beta E}{c}\right)c}{\gamma\left(E-\beta cp_3\right)-\gamma\left(p_3-\frac{\beta E}{c}\right)c}\right)}}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E-\beta cp_3\right)+\left(cp_3-\beta E\right)}{\left(E-\beta cp_3\right)-\left(cp_3-\beta E\right)}\right)}}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)-\beta\left(E+cp_3\right)}{\left(E- cp_3\right)+\beta\left(E- cp_3\right)}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\frac{1-\beta}{1+\beta}\right)}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\right)} + \frac{1}{2}\ln{\left(\frac{1-\beta}{1+\beta}\right)}}$$ $$y'=y - \frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$$

Hence, answer is (B)

5. The partition function of a single gas molecule is $Z_\alpha$. The partition function of $N$ such non-interacting gas molecules is given by
1. $\frac{(Z_\alpha)^N}{N!}$
2. $(Z_\alpha)^N$
3. $N(Z_\alpha)$
4. $\frac{(Z_\alpha)^N}{N}$

$Z_N=(Z_\alpha)^N$

Hence, answer is (B)