VLSI homework 1, and IoT reading assignment 1 done
This commit is contained in:
Aidan Sharpe
parent
3a16c8f0e9
commit
52f512b38c
@ -18,5 +18,14 @@
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@ -17,4 +17,26 @@
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||||
\tableofcontents
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||||
\pagebreak
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||||
|
||||
\chapter{}
|
||||
|
||||
$$\boxed{\text{Source}} \to \boxed{\text{Transmitter}} \to \boxed{\text{Channel}} \to \boxed{\text{Receiver}} \to \boxed{\text{Sink}}$$
|
||||
|
||||
\section{The Fundamental Transmission Limit}
|
||||
\thm{Shannon's Theorem}
|
||||
{
|
||||
The theoretical limit for error-free transmission in a communications system in the presence of noise (the channel capacity) is a function of the channel bandwidth $B$ and the signal to noise power ration $S/N$.
|
||||
\begin{equation}
|
||||
C = B \log_2(1 + S/N)
|
||||
\end{equation}
|
||||
}
|
||||
|
||||
\ex{Shannon's Theorem}
|
||||
{
|
||||
Given a 1[W] signal perturbed by 1[mW] of noise, the SNR is 1000. In dB, the SNR is
|
||||
\begin{equation}
|
||||
\text{dB} = 10 \log_{10}(\text{SNR}).
|
||||
\end{equation}
|
||||
In this case, the SNR is 30[dB].
|
||||
}
|
||||
|
||||
\end{document}
|
||||
|
||||
@ -1 +1,3 @@
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||||
\contentsline {chapter}{\numberline {1}}{2}{chapter.1}%
|
||||
\contentsline {section}{\numberline {1.1}The Fundamental Transmission Limit}{2}{section.1.1}%
|
||||
\contentsfinish
|
||||
|
||||
@ -1,14 +1,39 @@
|
||||
1. Who is leading the charge in IoT cybersecurity?
|
||||
All the organizations that helped publish the documents (GSMA, CISA, NSA, FBI, NCSC-UK, ACSC, CCCS, NCSC-NZ), and their references and acknowledgements (Microsoft, IBM, Nozomi Networks)
|
||||
|
||||
2. What are some of the positives of IoT?
|
||||
IoT allows more parts of our lives to be connected.
|
||||
|
||||
3. What are some of the negatives of IoT?
|
||||
By introducing so many more devices on our networks, attack surfaces are greatly increased.
|
||||
|
||||
4. How can IoT automation introduce new vulnerabilities into networks?
|
||||
IoT automation devices "directly impact human physical lives" (GSMA document). The sheer number of devices drastically increases the potential attack surface, and most devices lack an accessible and straightforward way to apply patches and firmware updates.
|
||||
|
||||
5. What are the challenges of IoT evolution?
|
||||
- The availability challenge: IoT devices are expected to have very high up-time
|
||||
- The identity challenge: devices must be able to accurately and securely identify themselves (think confidentiality and integrity)
|
||||
- The privacy challenge: since these devices integrate into our lives, privacy is paramount (think confidentiality)
|
||||
- The security challenge: ensuring that large amounts of users and physical systems are not exposed to risk (think confidentiality, integrity, and availability for IoT services and devices)
|
||||
|
||||
6. Which of the challenges is the largest hindrance to IoT developers?
|
||||
Security is a probably the largest challenge for IoT developers.
|
||||
|
||||
7. Which group, CISA or GSMA, are more industry specific, and which is more generic?
|
||||
The CISA focuses on cybersecurity and infrastructure security in general, while the GSMA focuses its efforts on mobile networks.
|
||||
9. What is/are the most important reference(s) in each document?
|
||||
|
||||
11. Within each document, when is the next update?
|
||||
12. What are the 2 key components that make up the IoT model, and which one does Rowan’s CS06440 - Cloud Computing and Internet of Things fall within?
|
||||
13. What is the goal of a risk assessment?
|
||||
14. Cite one example of a specific IoT device used in either document and what it was used to explain?
|
||||
8. What is/are the most important reference(s) in each document?
|
||||
The CISA document makes many references to supporting documents published by leading cycbersecurity agencies from around the world. These supporting documents are refrenced to provide the reader with further guidance in areas where they are focusing their individual efforts.
|
||||
|
||||
Since the purpose of the GSAM document is to act as an overview for more focused documents, it cites the supporting documents to provide supplemental guidance.
|
||||
|
||||
9. Within each document, when is the next update?
|
||||
|
||||
10. What are the 2 key components that make up the IoT model, and which one does Rowan’s CS06440 - Cloud Computing and Internet of Things fall within?
|
||||
The two key components are the service ecosystem and the endpoint ecosystem.
|
||||
|
||||
11. What is the goal of a risk assessment?
|
||||
Risk assessments are tests that attempt to quantify risk for a given system, how likely each vulnerability is to occur, and how much damage could be done if that vulnerability were to be exploited. Afterwards, the risk assessment may be used to determine which vulnerabilities are possible to resolve and which should be addressed first.
|
||||
|
||||
12. Cite one example of a specific IoT device used in either document and what it was used to explain?
|
||||
Digital door-locks are cited as an example of a common IoT endpoint in the GSMA document.
|
||||
|
||||
BIN
7th-Semester-Fall-2024/VLSI/homework/homework-1/Figure_1.png
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7th-Semester-Fall-2024/VLSI/homework/homework-1/Figure_1.png
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|
After Width: | Height: | Size: 48 KiB |
@ -0,0 +1,88 @@
|
||||
# VLSI Homework 1 - Aidan Sharpe
|
||||
|
||||
## Problem 1
|
||||
```python
|
||||
import numpy as np
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
e_0 = 8.85E-14
|
||||
k_ox = 3.9
|
||||
e_ox = e_0 * k_ox
|
||||
|
||||
mu = 350 # 350 cm^2 / (V*s)
|
||||
t_ox = 100E-8 # 100A
|
||||
|
||||
C_ox = e_ox / t_ox
|
||||
W = 1.2
|
||||
L = 0.6
|
||||
|
||||
beta = mu * C_ox * W/L
|
||||
|
||||
V_g_range = np.arange(6)
|
||||
V_t = 0.7
|
||||
V_dd = np.linspace(0, 5, 50)
|
||||
|
||||
def I_ds(V_gs, V_ds, V_t, beta):
|
||||
if V_gs < V_t:
|
||||
return np.zeros_like(V_ds)
|
||||
|
||||
V_dsat = V_gs - V_t
|
||||
|
||||
return np.where(V_ds < V_dsat,
|
||||
beta * (V_gs - V_t - V_ds/2) * V_ds,
|
||||
beta * (V_gs - V_t)**2 / 2)
|
||||
|
||||
for V_s in (0, 2):
|
||||
for V_g in V_g_range:
|
||||
V_gs = V_g - V_s
|
||||
V_ds = V_dd - V_s
|
||||
plt.plot(V_ds,
|
||||
1000*I_ds(V_gs, V_ds, V_t, beta),
|
||||
label="$V_{gs}$ = "+str(V_gs))
|
||||
|
||||
plt.xlabel("$V_{ds}$[V]")
|
||||
plt.ylabel("$I_{ds}$[mA]")
|
||||
plt.legend()
|
||||
plt.show()
|
||||
```
|
||||
|
||||
|
||||
## Problem 2
|
||||
Show that a MOSFET with channel length $2L$, when operating in the linear region, has the same current $I_{DS1}$ as two MOSFETs in series each with channel length $L$. Assume the same gate voltage $V_{DD}$ for all MOSFETs in the design and the same supply voltage $V_{DS}$ for both designs.
|
||||
|
||||
### Known
|
||||
The current through a MOSFET in the linear region is:
|
||||
$$I_{DS} = \beta\left(V_{GS} - V_t - \frac{V_{DS}}{2}\right)V_{DS}$$
|
||||
The parameter $\beta$ is defined as:
|
||||
$$\beta = \mu C_\text{ox} \frac{W}{L}$$
|
||||
|
||||
### Defining Variables
|
||||
The factor $\beta_1$ for the singular MOSFET is $\mu C_\text{ox} \frac{W}{2L}$, and the factor $\beta_2$ for each of the MOSFETs in series is $\mu C_\text{ox} \frac{W}{L}$. Therefore, $\beta_2 = 2\beta_1$.
|
||||
|
||||
The current $I_{DS1}$ through the singular MOSFET is:
|
||||
$$I_{DS1} = \beta_1\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS}$$
|
||||
The current $I_{DS2}$ through both MOSFETs in series is both:
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_1 - V_t - \frac{V_{DS} - V_1}{2}\right)(V_{DS} - V_1)$$
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
### Expanding $I_{DS2}$
|
||||
The first equation for $I_{DS2}$ can be rewritten as the difference of two terms:
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)(V_{DS} - V_1)$$
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
The term $-\frac{V_1 V_{DS}}{2}$ can be pulled out of the first term, and its complement can be pulled out of the second term. Therefore, $I_{DS2}$ can be simplified to the difference of two terms each in the form of the current through a MOSFET in the linear region.
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
### Reducing $I_{DS2}$
|
||||
Notice that the second term is the inverse of the second equation for $I_{DS2}$ under "Defining Variables". Setting them equal:
|
||||
$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_1}{2}\right)V_1 = \beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
Therefore:
|
||||
$$\beta_2\left(V_{DD} -V_t - \frac{V_{DS}}{2}\right)V_{DS} = 2\beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
Substituting $\beta_2$ on the left for $2\beta_1$ gives:
|
||||
$$2\beta_1\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} = 2\beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
|
||||
|
||||
Therefore:
|
||||
$$2I_{DS1} = 2I_{DS2}$$
|
||||
$$\boxed{I_{DS1} = I_{DS2}}$$
|
||||
BIN
7th-Semester-Fall-2024/VLSI/homework/homework-1/homework-1.pdf
Normal file
BIN
7th-Semester-Fall-2024/VLSI/homework/homework-1/homework-1.pdf
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42
7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_1.py
Normal file
42
7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_1.py
Normal file
@ -0,0 +1,42 @@
|
||||
import numpy as np
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
e_0 = 8.85E-14
|
||||
k_ox = 3.9
|
||||
e_ox = e_0 * k_ox
|
||||
|
||||
mu = 350 # 350 cm^2 / (V*s)
|
||||
t_ox = 100E-8 # 100A
|
||||
|
||||
C_ox = e_ox / t_ox
|
||||
W = 1.2
|
||||
L = 0.6
|
||||
|
||||
beta = mu * C_ox * W/L
|
||||
|
||||
V_g_range = np.arange(6)
|
||||
V_t = 0.7
|
||||
V_dd = np.linspace(0, 5, 50)
|
||||
|
||||
def I_ds(V_gs, V_ds, V_t, beta):
|
||||
if V_gs < V_t:
|
||||
return np.zeros_like(V_ds)
|
||||
|
||||
V_dsat = V_gs - V_t
|
||||
|
||||
return np.where(V_ds < V_dsat,
|
||||
beta * (V_gs - V_t - V_ds/2) * V_ds,
|
||||
beta * (V_gs - V_t)**2 / 2)
|
||||
|
||||
for V_s in (0, 2):
|
||||
for V_g in V_g_range:
|
||||
V_gs = V_g - V_s
|
||||
V_ds = V_dd - V_s
|
||||
plt.plot(V_ds,
|
||||
1000*I_ds(V_gs, V_ds, V_t, beta),
|
||||
label="$V_{gs}$ = "+str(V_gs))
|
||||
|
||||
plt.xlabel("$V_{ds}$[V]")
|
||||
plt.ylabel("$I_{ds}$[mA]")
|
||||
plt.legend()
|
||||
plt.show()
|
||||
29
7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_2.py
Normal file
29
7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_2.py
Normal file
@ -0,0 +1,29 @@
|
||||
import numpy as np
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
e_0 = 8.85E-14
|
||||
k_ox = 3.9
|
||||
e_ox = e_0 * k_ox
|
||||
|
||||
mu = 350 # 350 cm^2 / (V*s)
|
||||
t_ox = 100E-8 # 100A
|
||||
|
||||
C_ox = e_ox / t_ox
|
||||
W = 1.2
|
||||
L = 0.6
|
||||
|
||||
beta = mu * C_ox * W/L
|
||||
|
||||
V_t = 0.7
|
||||
|
||||
def I_ds(V_gs, V_ds, V_t, beta):
|
||||
if V_gs < V_t:
|
||||
return np.zeros_like(V_ds)
|
||||
|
||||
V_dsat = V_gs - V_t
|
||||
|
||||
return np.where(V_ds < V_dsat, beta * (V_gs - V_t - V_ds/2) * V_ds, beta * (V_gs - V_t)**2 / 2)
|
||||
|
||||
V_g = 0.8
|
||||
V_d1 = 1
|
||||
V_s1 =
|
||||
@ -1 +0,0 @@
|
||||
,sharpe,dhcp-150-250-209-16,09.09.2024 08:39,file:///home/sharpe/.var/app/org.libreoffice.LibreOffice/config/libreoffice/4;
|
||||
Binary file not shown.
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Reference in New Issue
Block a user