so together is the lab assignment 3 with Cadence and hw4 using Matlab

hw4_data_5_.txt

lab_assignment_3.pdf

hw4_7_.pdf

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Problem 1

{{ 0.814709, 0.179975, 0.551233, -0.291919},

{ 0.289709, 0.697138, -0.655795, 0.170119},

{-0.502313, 0.693979, 0.515825, 0.802216},

{ 0, 0, 0, 1 }}

Problem 2

{{-0.424721, 0.889030, -0.170987, 0.837701},

{ 0.754879, 0.452030, 0.475212, 0.205787},

{ 0.499769, 0.072758, -0.863097, 1.857961},

{ 0, 0, 0, 1 }}

Problem 3

{

t^2*Cos[t] – 3*Sin[t],

-(2*Sin[t]) – t*Cos[t],

t^2 + Sin[t] – 2*t*Cos[t]}

Problem 4

{{ 0.519014, 0.145713, 0.842254, -0.949724},

{-0.784234, 0.473125, 0.401409, 2.916842},

{-0.340002, -0.868861, 0.359832, -1.800944},

{ 0, 0, 0, 1 }}

Problem 5

{{-0.385475, 0.539666, 0.748444, 1.002016},

{ 0.453394, 0.817235, -0.355754, 1.390754},

{-0.803644, 0.202206, -0.559705, 1.519792},

{ 0, 0, 0, 1 }}

{{ 0.285402, 0.880153, -0.379311, 1.213720},

{ 0.710060, -0.460000, -0.533118, -3.012069},

{-0.643708, -0.117181, -0.756246, -1.815367},

{ 0, 0, 0, 1 }}

Lab Assignment 3

Introduction:

The purpose of this lab is to investigate inverter sizing techniques and minimize delay for driving

a large capacitive load, verify the theoretical concept. In this assignment, we will design an inverter

chain, calculate the propagation delay and find optimum number of stages to minimize delay.

1. Understanding impact of load on propagation delay:

a. Creating an Inverter:

i.

Create a new Cell-view. From virtuoso command window, click File → New →

Cell-view. In the popup window select the name of the library that you created

before. Give a name to the component, view name “schematic” and open with

“Schematic L”. Then click OK.

ii.

Following the previous lab, draw an inverter. Use NMOS_VTG and PMOS_VTG

as models. Set the parameters of MOSFETs as following:

PMOS: Length = 45n, Width = 90n

NMOS: Length = 45n, Width = 45n

This inverter will act as ‘unit inverter’ in the inverter chain.

Connect a Vpulse as the input of the inverter (Create → Instance, ‘Add instance’

window should open. Set Library → analogLib, cell → vpulse). Set the following

parameters for the source:

iii.

b. Creating Load

i.

Draw another inverter. Use NMOS_VTG and PMOS_VTG as models. Set the

parameters of MOSFETs as following:

PMOS: Length = 45n, Width = 90n x 64 = 5760n

NMOS: Length = 45n, Width = 45n x 64 = 2880n

With large gate width, this inverter will have a high gate capacitance. This gate

capacitance will act as load capacitance. Note that, size of the load inverter = 64 x unit

inverter

Connect the inverter with the previous one as shown in the figure:

ii.

iii.

iv.

Assign names to the wires that needs to viewed. Go to Create → Wire Name.

“Add Wire Name” window should appear. Set ‘Names’ to your preferences.

Measure the propagation delay. Note that, as two inverters are connected back

to back, output signal will follow input signal (If input raises to high, output

will also raise to high afterwards).

Now, increase the size of the load as followings and find the propagation delay

for both cases:

• PMOS: Length = 45n, Width = 90n x 32 = 2880n; NMOS: Length =

45n, Width = 45n x 32 = 1440n (size of the load inverter = 32 x unit

inverter)

• PMOS: Length = 45n, Width = 90n x 128 = 11520n; NMOS: Length =

45n, Width = 45n x 128 = 5760n (size of the load inverter = 128 x unit

inverter)

What change do you observe in propagation delay? Put an explanation for it.

For report, please add a plot on ‘Delay vs Gate Size of Load’ which

consists of all three cases. You can do it by hand, or you can use a simple

wave plotting tools (e.g: MATLAB).

2. Inverter sizing to minimize delay

i.

In this section, we will study the impact of additional inverters connected

in inverter chain and analyze the impact of sizing inverters while driving

large load.

ii.

Set the sizing at the load inverter as following:

PMOS: Length = 45n, Width = 90nx64 = 5760n

NMOS: Length = 45n, Width = 45nx64 = 2880n

iii.

Connect another inverter in-between unit inverter and load to create a

2-stage inverter chain. Keep the following sizing for the inverter:

PMOS: Length = 45n, Width = 90n

NMOS: Length = 45n, Width = 45n

Note that, newly connected inverter has the same size as unit inverter.

As, three inverter are connected back-to-back (look at the following figure),

output will inversely follow the input signal (if input raises to high, output

will fall to low and vice versa.).

iv.

To measure propagation delay, consider positive edge (Low → High)

of output signal. This edge is suggested for the sake of convenience.

v.

Now change the sizing of the newly connected inverter as follows and

measure the propagation delay:

PMOS: Length = 45n, Width = 720n

NMOS: Length = 45n, Width = 360n

What change do you observe in propagation delay? Provide an explanation for the

observed changes.

Now create three stage inverter chain. Set the up-sizing of the inverters for minimum delay with

3-stage inverter chain (follow class lecture. Find, upsizing factor for N=3). Measure the

propagation delay.

i.

For the sake of convenience, consider positive edge of output signal only. So for,

even numbers of inverter in inverter-chain measure following delay:

ii.

For odd numbers of inverters in inverter-chain, measure following delay:

iii.

Now create a 4-stage and 5-stage inverter chain. Set the up-sizing factors for

minimum delays in both cases. Measure propagation delay.

iv.

What change do you observe in propagation delay? Provide an explanation.

Complete the following table:

N

1

2

3

4

5

v.

U

Find the optimum number of stages required to minimize delay.

tp

Department of Electrical and Computer Engineering

University of Illinois at Chicago

ECE 452

Homework 4

Date: 3/6/2019

Due date: 3/12/2019

T

T

1. A rigid body is moving along a screw with the axis l = { 3 −1 2 +λ 3 5 −2.25

,λ ∈

R} with the pitch h = 3 and the magnitude θ(t) = t3 − 3t. At t = 0, the rigid body transformation between the body-fixed frame B and the global frame A is:

0.814709 0.179975

0.551233 −0.291919

0.289709 0.697138 −0.655795

0.170119

.

gab (0) =

−0.502313 0.693979

0.515825

0.802216

0

0

0

1

(a) Compute the transformation matrix gab (t) at t = 2s.

(b) Compute the spatial velocity Vabs (t).

(c) Compute the body velocity Vabb (t).

Note: Problem 1b (and to a certain extent 1c) would be appropriate for the exam, you should

be able to find the solution without the help of Matlab/Mathematica.

2. Consider a rigid body with the body fixed frame B, whose configuration with respect to the

frame A at time t0 is given by:

−0.424721 0.889030 −0.170987 0.837701

0.754879 0.452030

0.475212 0.205787

gab (t0 ) =

0.499769 0.072758 −0.863097 1.857961 .

0

0

0

1

The spatial velocity of the rigid body at time t0 (in vector form) is

Vabs (t0 ) =

−0.2 −1. 0.2 −0.8 0.6 0.4

T

.

Find a screw motion g(t) that passes through gab (t0 ) at time t0 and would result in the same

spatial rigid body velocity at t0 . Note that g(t) could be interpreted as the screw motion that

the rigid body follows instantaneously at t0 .

3. Consider two frames A and B that are initially aligned (the axes are parallel, but the origins

do not coincide). Assume that the frame B rotates with the rotational velocity described by

T

the vector −3.75 −3 4

in the frame A and that the origin of the frame B moves along

2

T

the curve t cos t − 3 sin t −t cos t − 2 sin t t2 − 2t cos t + sin t .

(a) Find the spatial velocity Vabs as a function of time.

(b) Find the body velocity Vabb as a function of time.

1

(c) Find the transformation gab at time t = 1s.

Note: Problem 3a (and to a certain extent Problems 3b and 3c) would be appropriate for the

exam, you should be able to find the solution without the help of Matlab/Mathematica.

4. The spatial velocity of the rigid body at t = 3s is equal to

Vabs (3) =

−0.6 1.8. 1.5 −2.1 1.5 0.3

T

,

(in the twist coordinates). What is the body velocity of the rigid body (in the vector form)

at t = 3s if you know that

0.519014

0.145713 0.842254 −0.949724

−0.784234

0.473125 0.401409

2.916842

gab (3) =

−0.340002 −0.868861 0.359832 −1.800944 ?

0

0

0

1

Note: Problem 4 does require more extensive computations and would be somewhat borderline for the exam. But I might use a similar problem if the rest of the problems require

minimal computation. So think how you could compute the solution without the help of Matlab/Mathematica.

5. Consider three frames, A, B, and C. The spatial velocity of frame B

T

is constant and equal to Vabs = 4 −1 2 −6 3 −7 . The body

relative to frame B is constant and equal to Vbcb = 1 −5 3 −2 −7

have:

−0.385475 0.539666

0.748444 1.002016

0.453394 0.817235 −0.355754 1.390754

gab (0) =

−0.803644 0.202206 −0.559705 1.519792

0

0

0

1

and

relative to frame A

velocity of frame C

T

9 . At t = 0, we

,

0.285402

0.880153 −0.379311

1.213720

0.710060 −0.460000 −0.533118 −3.012069

gbc (0) =

−0.643708 −0.117181 −0.756246 −1.815367 .

0

0

0

1

Find the following quantities:

(a) Write the formulas (but do not compute) for gab (t) and gbc (t).

(b) Compute the body velocity Vacb .

(c) Compute the spatial velocity Vacs .

Hint: A motion with a constant velocity corresponds to a screw motion. See also page 58 in

Murray, Li and Sastry.

2

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