In this post, we will see the step response of control system with and without PI controller.

**Step response without PI controller**

A unity feedback closed loop system has a open loop transfer function as shown below

We can see that the steady state error for a unit step input is 0.5 for the given system.

It is also seen that the response has settled at 0.5, so steady state error is 1-0.5=0.5 (Refer the response below)

**Step response with PI controller**

We saw in the video lecture that PI controller is used to decrease the steady state error without affecting the stability.

Let us see, how the PI controller denoted by will improve the steady state error of the system G(s) described above. So after adding PI controller in the forward path , overall closed loop looks like the diagram below, where (R(s) is the input i.e unit step function, since we are interested in step response)

Let be

i.e

So the combined open loop transfer function (OLTF) is =

After unit step input is applied as shown in the block diagram, the unit step response looks like below.

So , we can see the difference in step response before and after adding Proportional Integral controller(PI). PI controller is used to improve steady state response in cases like these. Always remember that increasing the type of the system decreases the steady state error.

In this case, introducing PI controller in the forward path has helped us achieve zero steady state error for step input instead of 0.5. If you have any doubts regarding this topic, please use comments section.

]]>G(s)H(s) = lies

a) Between -2 and origin

b) Between -2 and -5

c) Between -10 and -infinity

d) At infinity

2) A unity feedback system has open loop transfer function G(s)H(s)= . Its root locus plot intersects the jw axis at

a)

b)

c)

d) Does not intersect the jw axis

3) A unity feedback system has open loop transfer function G(s)H(s)= . What is the maximum value of K for which the given open loop TF is stable enough?

a)

b) 6

c) 10

d) 6

4) The value of K for which the unity feedback system

= crosses the imaginary axis is

a) 8

b) 16

c) 48

d)64

…….

** BODE PLOT QUESTIONS***

5) The Bode plot for the gain magnitude of a minimum phase system G(s) is shown in the figure. The transfer function G(s) is

a)

b)

c)

d)

6) For a transfer function . Maximum phase lead of the compensator is (GATE EE-2000)

a) 52 deg at 4 rad/s

b) 52 deg at 10 rad/s

c) 55 deg at 12 rad/s

d) None of the above.

]]>

a) Silicon acts as p-type dopants in Arsenic sites and n-type dopants in Gallium sites

b) Silicon acts as n-type dopants in Arsenic sites and p-type dopants in Gallium sites

c) Silicon acts as p-type dopants in both Arsenic and Gallium sites

d) Silicon acts as n-type dopants in both Arsenic and Gallium sites

Correct Ans : a

Background: amphoteric dopant -> element which can act either as a donor or an acceptor in a given semiconductor; e.g. Si is an amphoteric dopant of GaAs where it acts as a donor on a Ga site or as an acceptor on an As site.

2)

Correct Answer : B

Hint : Apply Kirchoff’s law for the branch involving 10k, 5k and 3v source with vin =15v and for vin=-15v

3) A good transconductance amplifier should have

a) High input resistance and low output resistance

b) Low input resistance and high output resistance

c) High input and output resistance

d) Low input and output resistance

Correct Ans : c

4) Which one of the following is the general solution of this first order differential equation

= where x and y are real.

a) y=1+x +

b) y=1+x+tan(x+c)

c) y=1-x+

d) y=1-x+tan(x+c)

where c is a constant

Correct Answer : d

5)

Correct Ans : 43.3 to 45.3

6) Assuming that transistors M1 and M2 are identical and have a threshold voltage of 1V, the state of transistors M1 and M2 are respectively

a) saturation, saturation

b) Linear, linear

c) saturation,linear

d) Linear, saturation

For MOSFET , the condition of saturation is

>=

Here = 3V, =2.5V, =2V(Since source is 0 for M1) =1V

so >=

>= =2.5-1 = 1.5V

Since = 3V > 1.5V, M2 is in saturation.

Assuming M1 is also in saturation, then same drain current flows in both M1 and M2. Since the both MOSFET’s are identical , we can write

=

=

2-1= 2.5 – -1

1=1.5 –( which is nothing but drain of M1)

=0.5V, which is less than , which is equal to 1V

0.5V < 1V, so the M1 transistor is not in saturation, but in Linear region. Hence the correct answer is “d”

7) Standard air filled rectangular waveguides of dimensions a=2.29cm and b=1.02cm are designed for radar applications. It is desired that these waveguides operate only in the dominant mode with the operating frequency at least 25% above the cut-off frequency of the mode, but not higher than the 95% of the next higher cut- off frequency. The range of the allowable frequency is

Note (<= refers to lesser than or equal to)

a) 8.19 GHz<= f <=13.1 GHZ

b) 8.19 GHz <= f <=12.45 GHZ

c) 6.155 GHz<= f <=13.1 GHZ

d) 1.64 GHz<= f <=10.24 GHZ

8)

Solution : Since of the transistor is unity, it means , we can assume

Voltage at the base of the transistor = 4.8V

=0.8V

so,

4.8-

So, =4/2k = 2 mA , which means =2 mA.

Applying Kirchoff’s law to the transistor on collector and emitter side

=18 -12=6V

9) Consider a characteristic equation given by

, the condition for stability is

a) k > 5

b) -10 < k

c) k > -4

d) -10 < k < -4

10) For a unit step input, a system with a closed loop transfer function of

has a steady state output of

a) 10

b) 5

c) 2

d) 4

11) For the system , the approximate time taken for a step response to reach 98% of its final value is

a) 1 s

b) 2 s

c) 4 s

d) 8 s

]]>

a) Real, but not equal

b) Real and equal

c) Complex conjugates

d) Imaginary

2) For a second order control system with the closed loop transfer function

T(s) = , the settling time for 2 percent band is (In seconds)

a) 1.5

b)2

c) 3

d) 4

3) Consider a system with transfer function G(s)= . Its damping ratio will be 0.5 when the value of K is

a) 2/6

b) 3

c) 1/6

d) 6

4) A unity feedback control system has the open loop transfer function

G(s)= . If the input to the system is a unit ramp, the steady state error will be

a) 0

b) 0.5

c) 2

d) Infinity

5) A causal system having the transfer function G(s) = is excited with 10 u(t). The time at which the output reaches 99% of its steady state value is

a) 2.7 seconds

b) 2.5 seconds

c) 2.3 seconds

d) 2.1 seconds

6) A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are

a) 1 and 20

b) 0 and 20

c) 0 and 1/20

d) 1 and 1/20

]]>

a) Increases

b) Decreases

c) Remains same

d) Zero

2) Air filled waveguide has a dominant mode cut off frequency of 9 GHz. One of the dimensions of the waveguide is

a) 4.3 cm

b) 1.66 cm

c) 3.3 cm

d) 0.8 cm

3) A 50 ohm characteristic impedance line is connected to load which has a reflection coefficient of 0.268. If vin=15v, net power delivered to load will be

a)0.139 W

b)1.39 W

c)0.278 W

d)2.78 W

4) A waveguide has a separation of 3 cm for the broader dimension and carries the dominant mode at an unknown frequency. If the wave impedance is 550 ohm, the unknown frequency f is

a) 7.66 GHz

b) 8.66 GHz

c) 6.66 GHz

d) 10.66 GHz

]]>= 2 and

= 2

the wave is

a) Linearly polarized

b) Elliptically polarized

c) Left circularly polarized

d) Right circularly polarized

2) The electric field of an electromagnetic wave propagating in the positive z direction is

= – + – The wave is

a) Linearly polarized in the Z direction

b) Elliptically polarized

c) Left Hand circularly polarized

d) Right Hand circularly polarized

3) The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction, is given by

```
E =
The frequency and polarization of the wave are
a) 1.2 GHz and left circular
b) 4 GHz and left circular
c) 1.2 GHz and right circular
d) 4 GHz and right circular
4) A plane wave propagating in the dielectric medium has an electric field given as
E
```_{x } = E_{0}Cos (2.6×10^{10}t -100z). The phase velocity of plane wave is:

a)

b)

c)

d)

]]>

1) B=

2) B=

3) B=

4) B=

Correct Answer : 2

]]>a) along jw axis in s-plane

b) on a rectangular strip in s-plane

c) along a circle in s-plane

d) on a circular strip in s-plane

**2) Convolution of two voltage pulses of amplitude 2 Volt and width 2 sec is a
**

a) Rectangular pulse

b) square pulse

c) Trapezoidal

d) Triangular pulse

**3) The FT of v(t) is for |w|<=1. The Energy dissipated by 1 resistor, if v(t) is applied to it is**

a) 1/

b) 2/

c)

d)

**4) equals**

a)2

b) 2 u(t)

c)

d) 0.5

5) A box contains 12 balls numbered from 1 to 12. If a ball is taken at random, what is the probability of getting a ball with a number which is a multiple of either 2 or 3.

______________

6) A box contains 2000 transistors of which 5 % are defective. A second box contains 500 transistors of which 40% are defective. Two other boxes contain 1000 transistors of which 10% are defective transistors. A single transistor is taken at random from one of the boxes.

i) What is the probability that selected transistor is defective?

ii) What is the probability that defective transistor is taken from box 2?

__________

……………………………………

Post your answers in the comment section

1-a 2-d 4-d

6 – Use Baye’s theorem

]]>a) Changes by a factor AB

b) Decreases by a factor 1+AB

c) Increases by a factor 1+AB

d) Decreases by a factor A/(1+AB)

2) Negative feedback in voltage series Amplifier

a) Increases input impedance and decreases output impedance

b) Decreases input impedance and decreases output impedance

c) Decreases input impedance and increases output impedance

d) Increases input impedance and increases output impedance

**3) Emitter follower is a **

a) Current series feedback amplifier

b) Voltage series feedback amplifier

c) Current shunt feedback Amplifier

d) Voltage shunt feedback amplifier

**4) Which of the following parameters won’t get affected by feedback?**

a) Gain

b) Bandwidth

c) Gain Bandwidth product

d) None of these

**5) Effect of feedback on output Offset voltage of an OPAMPS**

a) Increases

b) decreases

c) Remains same

d) None of these

All have their advantages and disadvantages. However, in the past decade, CMOS is ruling the electronics field because of its SIZE and POWER.

CMOS(Complementary metal oxide semiconductor) has ZERO static power dissipation. However, The switching speed of CMOS is very low compared to BJT, and also CMOS cannot drive large current to the load. In order to drive large current to the load, the size of MOS needs to be significantly increased.

In order to incorporate the advantages of both CMOS and BJT, new logical family came into existence and that logical family is called BICMOS. In this post we will see the working of BICMOS NAND GATE and appreciate the work.

This is the two input BICMOS NAND gate.

It consists of

1) 2 PMOS PA and PB

2) 4 NMOS, NA1, NB1, NA3, NB3

3) Two BJT’s QP and Q0

NAND GATE

A B Y

0 0 1

0 1 1

1 0 1

1 1 0

**Purpose of Various devices**

- PA, PB, NA1 and NB1 are used for logical purposes(Just like in CMOS)
- N2, NB3 and NA3 are used to remove base charge from the transistors.

**Why to remove the base charge anyway?**

- For high switching speeds of the BJT’s, we need to remove the base charge from the transistor. In order to remove the charge from the base of the transistor we need a mechanism, that can be achieved by using these 3 NMOS.
- If we do not remove the base charge, the transistor will be in ON state and takes a lot of time to go to the OFF state. If we do not remove the base charge, the half purpose of BICMOS will be lost.

**Case 1 : Both A and B are low.**

If both the inputs VA and VB(Refer the circuit) are low(0). PA and PB will be ON and the base of QP will be high. Thus the top BJT(Q) will be ON which pulls the output UP . If there is a capacitive load. The output current will be almost 101 times the base current i.e it will be 101*IB(Assuming beta of transistor is 100). (if there was no BJT, the output current would be just IB).

Also, note that N2 is getting input from the base of QA transistor. Since the base of QP is HIGH(say 5v), it will turn on N2 . Since N2 is ON, it will *PULL THE BASE CHARGE OUT OF Q0 transistor*. So, if both the inputs are low, QP is ON and Q0 is OFF. It is very important to appreciate the role of N2 here which facilitates high switching speed.

**Case 2 :Both A and B are high**

NB3 and NA3 NMOS make sure that QP is in OFF state. PA and PB are OFF. NB1, NA1 will be ON. The output is discharged via NB1, NA1 and Q0(High Speed). N2 will be effectively out of circuit in this case. So, Q0 is ON and QP is OFF.

I have discussed the 2 cases. You can correlate with the other 2 inputs.

**Disadvantages of BICMOS**

1) Fabrication cost is high

2) Due to VBE(Base to emitter voltage, 0.7v approximately) of BJT’s , desirable performance is not obtained when the BICMOS gates are operated at lesser voltages (3V, 2.4V) etc.