VLSI homework 10, ECOMMS homework 3

7th-Semester-Fall-2024/ECOMMS/homework/homework-3/homework-3.md

@@ -0,0 +1,95 @@
+# ECOMMS Homework 3 - Aidan Sharpe
+
+## Problem 1
+A sinusoidal signal $m(t) = \cos(2\pi f_m t)$ is the input to an angle-modulated transmitter, $A_c = 1$, and the carrier frequency is $f_c = 1$ Hz and $f_m = f_c/4$. Plot $m(t)$, and the corresponding phase and frequency modulated signals, $S_p(t)$ and $S_f(t)$ respecively. $D_p = \pi$ and $D_f = \pi$.
+
+```python
+f_c = 1
+f_m = f_c/4
+
+f_s = 8E3
+T_s = 1/f_s
+
+A_c = 1
+D_p = np.pi
+D_f = np.pi
+
+omega_m = 2*np.pi*f_m
+omega_c = 2*np.pi*f_c
+
+# Time samples from to 10 seconds with a sampling frequency of 8kHz.
+t = np.arange(0,10,T_s)
+
+# The message signal
+m = np.cos(omega_m * t)
+
+# The phase modulated signal
+S_p = np.cos(omega_c*t + D_p*m)
+
+# The time integral of the mesage signal
+M = np.sin(omega_m * t) / omega_m
+
+# The frequency modulated signal
+S_f = np.cos(omega_c*t + D_f*M)
+```
+
+![](1.png)
+
+## Problem 2
+An FM signal has sinusoidal modulation with a frequency of $f_m = 15$kHz and a modulation index of $\beta = 2.0$. Find the transmission bandwidth by using Carson's rule, and find the percentage of total FM signal power that lies within the Carson rule bandwidth.
+
+```python
+# Modulation index
+beta = 2.0
+
+# Message frequency
+f_m = 15E+3
+
+# Transmission bandwidth
+B_T = 2*(beta+1)*f_m
+```
+
+$\boxed{B_T = 90\text{kHz}}$
+
+```python
+A_c = 1
+n = np.arange(-3,4,1)
+
+# Evaluate the Bessel function at values in Carson rule bandwidth
+bessel_values = np.abs(sp.special.jv(n,beta))
+
+P_C = 0.5 * A_c**2 * np.sum(bessel_values**2))
+P = 0.5 * A_c**2
+```
+
+\boxed{P_c = 0.9976 P}
+
+## Problem 3
+A modulated RF waveform is given by $500\cos(\omega_c t + 20\cos(\omega_1 t))$, where $\omega_1 = 2\pi f_1$, $f_1 = 1$kHz, $\omega_c = 2\pi f_c$, and $f_c = 100$MHz.
+
+### 3a
+If the phase sensitivity $D_p = 100$ rad/V, find the mathematical expression for the corresponding phase modulation voltage $m(t)$. What is its peak value and frequency?
+
+$m(t) = \frac{20\cos(\omega_1 t)}{D_p} = 0.2\cos(\omega_1 t)$
+
+Peak value: $2 \times 10^{-1}$
+
+Frequency: 1kHz
+
+### 3b
+If the frequency deviation constant $D_f = 10^6$rad/Vs, find the mathematical expression for the corresponding FM voltage $m(t)$. What is its peak value and its frequency?
+
+$\theta(t) = 20\cos(\omega_1 t)$
+
+$m(t) = \frac{1}{D_f} \frac{d \theta(t)}{dt} = -2 \times 10^{-5} \omega_1 \sin(\omega_1 t)$
+
+Peak value: $2 \times 10^{-5}$
+
+Frequency: 1kHz
+
+### 3c
+If the RF waveform appears across a 50$\Omega$ load, determine the average power and the PEP.
+
+The average power of the signal is:
+$$\langle s^2(t) \rangle = \frac{1}{T} \int\limits_{-T/2}^{T/2} 500\cos(\omega_c t + 20\cos(\omega_1 t)) dt$$
+

7th-Semester-Fall-2024/ECOMMS/homework/homework-3/problem-1.py

@@ -4,21 +4,41 @@ import matplotlib.pyplot as plt
 
 f_c = 1
 f_m = f_c/4
+f_s = 8E3
+T_s = 1/f_s
+
+A_c = 1
 
 omega_m = 2*np.pi*f_m
 omega_c = 2*np.pi*f_c
 
-T_c = f_c
+# Time samples to 10 seconds at a samplig frequency of 8kHz
-
+t = np.arange(0,10,T_s)
-t = np.arange(0,10,T_c/10)
 m = np.cos(omega_m * t)
 
-plt.plot(t, m)
+plt.subplot(311)
-plt.show()
+plt.plot(t, m, label="$m(t)$")
+plt.ylabel("$m(t)$")
 
 D_p = np.pi
 D_f = np.pi
 
-S_p = np.cos(omega_c + D_p*m)
+# The phase modulated signal
+S_p = np.cos(omega_c*t + D_p*m)
+
+plt.subplot(312)
+plt.ylabel("$S_p(t)$")
+plt.xlabel("t (seconds)")
 plt.plot(t,S_p)
+
+# The time integral of the mesage signal
+M = np.sin(omega_m * t) / omega_m
+
+# The frequency modulated signal
+S_f = np.cos(omega_c*t + D_f*M)
+
+plt.subplot(414)
+plt.ylabel("$S_f(t)$")
+plt.xlabel("t (seconds)")
+plt.plot(t,S_f)
 plt.show()

7th-Semester-Fall-2024/ECOMMS/homework/homework-3/problem-2.py

@@ -4,6 +4,7 @@ import matplotlib.pyplot as plt
 
 # Modulation index
 beta = 2.0
+A_c = 1
 
 # Number of impulses is 2*(beta+1) + 1.
 # For beta=2, the number of impulses is 3 either side of the center frequency
@@ -15,12 +16,20 @@ f_m = 15E+3
 # Transmission bandwidth
 B_T = 2*(beta+1)*f_m
 
-n = np.arange(0,10,1)
+print(B_T)
-bessel_values = sp.special.jv(n,beta)
+
-bessel_power = np.cumsum(bessel_values)
+n = np.arange(-3,4,1)
+bessel_values = np.abs(sp.special.jv(n,beta))
+
+N = np.arange(-100,100)
+bessel_power = np.sum(np.abs(sp.special.jv(N,beta)))
 plt.stem(n, bessel_values)
-plt.plot(n, bessel_power)
+plt.plot(n, np.cumsum(bessel_values**2), label="Cumulative power in Carson rule bandwidth", color="red")
-plt.hlines(0.98*bessel_power[-1], xmin=n[0], xmax=n[-1])
+#plt.plot(n, bessel_power)
+plt.hlines(0.98, xmin=n[0], xmax=n[-1], label="98% of total power", color="orange")
+plt.legend()
 plt.show()
 
-# 
+P_C = 0.5 * A_c**2 * np.sum(bessel_values**2)
+P = 0.5 * A_c**2
+print("Percent of total power:", P_C / P)

7th-Semester-Fall-2024/VLSI/homework/homework-10/homework-10.md

@@ -0,0 +1,23 @@
+# VLSI Homework 10 - Aidan Sharpe
+
+## Problem 1
+You are synthesizing a chip composed of random logic with an average activity factor of $\alpha = 0.1$. You are using a standard cell process with an average switching capacitance of $C_\text{avg} = 450$[pF/mm$^2$]. Estimate the dynamic power consumption of your chip if it has an area of 70[mm$^2$] and runs at 450[MHz] at $V_{DD} = 0.9$[V].
+
+$P_\text{dynamic} = C_\text{sw} V_{DD}^2 f_\text{sw}$
+
+```python
+alpha = 0.1                     # Activity factor
+C_avg = 450E-12                 # Average switching capacitance [F / mm^2]
+A = 70                          # Chip area [mm^2]
+f = 450E+6                      # Clock frequency [Hz]
+V_DD = 0.9                      # Supply voltage [V]
+
+C_sw = C_avg * A                # Total switching capacitance [F]
+f_sw = alpha * f                # Switching frequency [Hz]
+
+P_sw = C_sw * V_DD**2 * f_sw    # Dynamic power [W]
+
+print(f"Dynamic Power: {P_sw}")
+```
+
+$\boxed{P_\text{dynamic} = 1.148175\text{[W]}}$

7th-Semester-Fall-2024/VLSI/homework/homework-10/homework-10.py

@@ -0,0 +1,12 @@
+alpha = 0.1                     # Activity factor
+C_avg = 450E-12                 # Average switching capacitance [F / mm^2]
+A = 70                          # Chip area [mm^2]
+f = 450E+6                      # Clock frequency [Hz]
+V_DD = 0.9                      # Supply voltage [V]
+
+C_sw = C_avg * A                # Total switching capacitance [F]
+f_sw = alpha * f                # Switching frequency [Hz]
+
+P_sw = C_sw * V_DD**2 * f_sw    # Dynamic power [W]
+
+print(f"Dynamic Power: {P_sw}")