6-transient Analysis
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ELECTRIC CIRCUITS THEORY 1 These lecture slides have been compiled by Mohammed LECTURE 6 SalahUdDin Ayubi. Transient Analysis 21 August 2005
Engineer M S Ayubi
1
Dynamic Circuit Analysis Two Approaches to solving circuits with energy storage elements and time-varying inputs: • Transient Analysis (time-domain) • Phasor Analysis (frequency-domain)
21 August 2005
Engineer M S Ayubi
2
Test Circuit
21 August 2005
Engineer M S Ayubi
3
Measured in Lab 6.00
i(t) (Amps)
5.00 4.00 3.00 2.00 1.00 0.00 0
2
4
6
8
10
Time (sec)
21 August 2005
Engineer M S Ayubi
4
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
5
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
6
First Order Transient Response • One (equivalent) energy storage element • Looking for V or I vs. time • Switch in the circuit creates a transient event at t=0 • Eventually reaches steady state
21 August 2005
Engineer M S Ayubi
7
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
8
RL Circuits
21 August 2005
Engineer M S Ayubi
9
RL Circuits
t=0- (just before) iL=0
t=0+ (just after) iL=0
Initial Condition
21 August 2005
t=∞ (long after) iL=Vs/R Final Condition
Engineer M S Ayubi
10
RL Circuits
KVL loop:
21 August 2005
Engineer M S Ayubi
11
RL Circuits
21 August 2005
Engineer M S Ayubi
12
RL Circuits
What happens at t=0? At t=∞?
21 August 2005
Engineer M S Ayubi
13
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form
21 August 2005
Engineer M S Ayubi
14
RC Circuits
21 August 2005
Engineer M S Ayubi
15
RC Circuits
t=0- (just before) VC=0
21 August 2005
t=0+ (just after) VC=0
Engineer M S Ayubi
t=∞ (long after) VC=IsR
16
RC Circuits KCL node: Solve…
21 August 2005
Engineer M S Ayubi
17
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
18
General Approach For 0 < t < ∞ Unknown = Final + (Initial – Final)e-t/τ τ = time constant Try t=0, t=∞ 21 August 2005
Engineer M S Ayubi
19
Initial Value VC or IL at t=0+ is same as value at t=0• This is the only value that is guaranteed to remain constant before and after the switch changes. • Assume circuit has remained in same state for a long time leading up to time of switch change. (Capacitor -> open, Inductor->short). Compute VC or IL using simplified circuit at t=021 August 2005
Engineer M S Ayubi
20
Final Value • Assume circuit has remained in same state for a long time after switch change. (Capacitor -> open, Inductor->short). Compute VC or IL using simplified circuit at t=∞
21 August 2005
Engineer M S Ayubi
21
Time Constant ∀ τ = RC ∀ τ = L/R Units are seconds
21 August 2005
Engineer M S Ayubi
22
General Approach For 0 < t < ∞
Steady State
Transient
Unknown = Final + (Initial – Final)e-t/τ
21 August 2005
Engineer M S Ayubi
23
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
24
Example 1 • Find i(t) through 2 ohm resistor
21 August 2005
Engineer M S Ayubi
25
Example 1: Initial At t=0-, inductor looks like short, but current is zero. i(t= 0-)=0
21 August 2005
Engineer M S Ayubi
26
Example 1: Initial At t=0+, inductor forces current to stay constant i(t= 0+)= i(t= 0-)=0 Initial = 0 A
21 August 2005
Engineer M S Ayubi
27
Example 1: Final At t=∞, inductor again looks like short i(t=∞)=10V / 2Ω = 5A Final = 5A
21 August 2005
Engineer M S Ayubi
28
Example 1: τ τ =L/R RTH seen by inductor is 2 Ω τ = 5H/2 Ω = 2.5 sec
21 August 2005
Engineer M S Ayubi
29
Example 1 i(t) = 5 + (0-5)e-t/τ 6.00
i(t) (Amps)
5.00 4.00 3.00 2.00 1.00 0.00 0
2
4
6
8
10
Time (sec)
21 August 2005
Engineer M S Ayubi
30
Time Constant 120.0%
1-e^-t/tau
100.0% 80.0%
86.5%
95.0%
98.2%
99.3%
99.8%
99.9%
3
4
5
6
7
63.2%
60.0% 40.0% 20.0% 0.0% 1
2
Multiple of tau
21 August 2005
Engineer M S Ayubi
31
Example 2 Find i(t) in left resistor
21 August 2005
Engineer M S Ayubi
32
Example 2: Initial i(t=0-)=10V/(2Ω || 2Ω ) = 10A i(t=0+)= i(t=0-)=10A
Note that current through resistor is 5A at t=0and 10A at t=0+ 21 August 2005
Engineer M S Ayubi
33
Example 2: Final i(t=∞)= 10V / 2Ω = 5A
21 August 2005
Engineer M S Ayubi
34
Example 2: Tau RTH seen by inductor during 0 < t < ∞ is 2 Ω τ = L/R = 2.5 sec
21 August 2005
Engineer M S Ayubi
35
Example 2 i(t) = 5 + (10 – 5)e-t/τ τ = 2.5 sec 11
i(t) (amps)
9
7
5
3 -1
1
3
5
7
9
11
Time (sec)
21 August 2005
Engineer M S Ayubi
36
Example 3 • Find current in left resistor
21 August 2005
Engineer M S Ayubi
37
Game Plan • Part A: Solve problem for 0 < t < 10s – Initial condition? – Final condition?
• Part B: Solve problem for 10 < t < ∞ – Initial condition? – Final condition?
21 August 2005
Engineer M S Ayubi
38
A: Initial Condition • Capacitor looks like open before t=0 • VC(0-)=5A ( 2Ω || 2Ω ) = 5V
21 August 2005
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39
A: Initial Condition • VC(0+)= VC(0-)=5V • iR(0+)=5V/2Ω =2.5A
21 August 2005
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40
A: Final Condition iR(∞)=5A VC(∞)=10V
21 August 2005
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41
A: Tau RTH seen by capacitor = 2 Ω τ = RC = (2Ω ) (3F) = 6 sec
21 August 2005
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42
A: Answer For 0 < t < 10 sec iR(t)=5 + (2.5 – 5)e-t/τ τ = 6 sec VC(t)=10 + (5-10) e-t/τ
21 August 2005
Engineer M S Ayubi
43
B: Initial Condition • Initial condition found from previous answer (not steady state) • VC(10-)=10 + (5-10) e-10/τ =9.06V • VC(10+)= VC(10-)=9.06V • iR(10+)=9.1V/2Ω = 4.53V
21 August 2005
Engineer M S Ayubi
44
B: Final Condition iR(∞)=2.5A
21 August 2005
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45
B: Tau RTH seen by capacitor = 2Ω || 2Ω = 1Ω τ = RC = (1Ω ) (3F) = 3 sec
21 August 2005
Engineer M S Ayubi
46
B: Answer for 10 < t < ∞ iR(t)=2.5 + (4.53 – 2.5)e-(t-10)/τ τ = 3 sec
21 August 2005
Engineer M S Ayubi
47
Example 3 Graph 5 4.5
i(t)
4 3.5 3 2.5 2 -5
0
5
10
15
20
Time (sec)
21 August 2005
Engineer M S Ayubi
48
Example 4 • Find the voltage across the capacitor Vc(t) for a square wave input (magnitude=10V, period=2ms)
21 August 2005
Engineer M S Ayubi
49
Example 4 • The initial capacitor voltage at t=0 is 0V, heading toward final voltage of 10V (but it will not reach final value before input waveform switches). • The capacitor voltage at t=1ms (which will be between 5 and 10 V) will be the new initial voltage, heading toward a final voltage of 0V (but it will not reach final value before input switches again). 21 August 2005
Engineer M S Ayubi
50
Example 4: Tau τ = RC = (1 KΩ )(1 uF) = 1ms So voltage will get a little over half way each time before the square wave input switches.
21 August 2005
Engineer M S Ayubi
51
Example 4 Pspice Run
21 August 2005
Engineer M S Ayubi
52
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
53
Second Order (RLC)
Series
21 August 2005
Parallel
Engineer M S Ayubi
54
Series RLC
21 August 2005
Engineer M S Ayubi
55
Parallel RLC
21 August 2005
Engineer M S Ayubi
56
General Form 2
d x(t ) dx(t ) a +b + cx(t ) = f (t ) 2 dt dt a b
Series 1 Rth/L
Parallel 1 1/(RthC)
c
1/(LC)
1/(LC)
21 August 2005
Engineer M S Ayubi
57
General Form Steady State
Transient
unknown = final value + natural response
21 August 2005
Engineer M S Ayubi
58
Final Value • Steady state response is considered constant, so use short-cut (capacitor -> open, inductor -> short) to find final value
21 August 2005
Engineer M S Ayubi
59
Natural Response • Guess that transient response is of the form: xtrans(t) = Aept • If constants A and p can be found, then this is the solution
21 August 2005
Engineer M S Ayubi
60
Natural Response 2
d x(t ) dx(t ) a +b + cx(t ) = f (t ) 2 dt dt
ap Ae + bpAe + cAe = 0 2
( ap
pt
pt
2
+ bp + c Ae = 0
)
pt
pt
ap + bp + c = 0 2
21 August 2005
Engineer M S Ayubi
61
Natural Response − b ± b − 4ac p1 , p2 = 2a 2
a, b, and c are circuit parameters. Roots p1, p2 are computed from these parameters. Three possibilities, depending on b2 compared to 4ac 21 August 2005
Engineer M S Ayubi
62
Real Distinct Roots b2 > 4ac Over Damped
xtrans (t ) = A1e
− p1t
+ A2 e
− p2t
Know p1 and p2, must find A1 and A2 from specific solution (using forcing function) 21 August 2005
Engineer M S Ayubi
63
Real Equal Roots b2 = 4ac Critically Damped
xtrans (t ) = A1e
− pt
+ A2te
− pt
Know p= p1=p2, must find A1 and A2 from specific solution (using forcing function)
21 August 2005
Engineer M S Ayubi
64
Complex Roots b2 < 4ac Under Damped
p1, 2 = −α ± jβ xtrans (t ) = Ce
−αt
sin( βt + φ )
Know α and β , must find C and φ from specific solution (using forcing function) 21 August 2005
Engineer M S Ayubi
65
Second Order Response 4 3.5 3
Real Distinct (Over Damped)
2.5 2
Real Equal (Critically Damped)
1.5 1
Complex (Under Damped)
0.5 0 0
10
21 August 2005
20
30
Engineer M S Ayubi
66
Phasor vs. Transient Analysis • Phasor Analysis – source is sine or cosine – no switches
• Transient Analysis – switch and source is constant – square wave can also be treated as a transient
21 August 2005
Engineer M S Ayubi
67
Transient Response • Many other physical systems react with the same type of response (at least approximately) – a first order exponential response starting at an initial condition, moving towards a final condition, with a time constant τ (e.g., temperature probe responding to instantaneous change in temperature)
21 August 2005
Engineer M S Ayubi
68
Engineer M S Ayubi
1
Dynamic Circuit Analysis Two Approaches to solving circuits with energy storage elements and time-varying inputs: • Transient Analysis (time-domain) • Phasor Analysis (frequency-domain)
21 August 2005
Engineer M S Ayubi
2
Test Circuit
21 August 2005
Engineer M S Ayubi
3
Measured in Lab 6.00
i(t) (Amps)
5.00 4.00 3.00 2.00 1.00 0.00 0
2
4
6
8
10
Time (sec)
21 August 2005
Engineer M S Ayubi
4
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
5
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
6
First Order Transient Response • One (equivalent) energy storage element • Looking for V or I vs. time • Switch in the circuit creates a transient event at t=0 • Eventually reaches steady state
21 August 2005
Engineer M S Ayubi
7
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
8
RL Circuits
21 August 2005
Engineer M S Ayubi
9
RL Circuits
t=0- (just before) iL=0
t=0+ (just after) iL=0
Initial Condition
21 August 2005
t=∞ (long after) iL=Vs/R Final Condition
Engineer M S Ayubi
10
RL Circuits
KVL loop:
21 August 2005
Engineer M S Ayubi
11
RL Circuits
21 August 2005
Engineer M S Ayubi
12
RL Circuits
What happens at t=0? At t=∞?
21 August 2005
Engineer M S Ayubi
13
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form
21 August 2005
Engineer M S Ayubi
14
RC Circuits
21 August 2005
Engineer M S Ayubi
15
RC Circuits
t=0- (just before) VC=0
21 August 2005
t=0+ (just after) VC=0
Engineer M S Ayubi
t=∞ (long after) VC=IsR
16
RC Circuits KCL node: Solve…
21 August 2005
Engineer M S Ayubi
17
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
18
General Approach For 0 < t < ∞ Unknown = Final + (Initial – Final)e-t/τ τ = time constant Try t=0, t=∞ 21 August 2005
Engineer M S Ayubi
19
Initial Value VC or IL at t=0+ is same as value at t=0• This is the only value that is guaranteed to remain constant before and after the switch changes. • Assume circuit has remained in same state for a long time leading up to time of switch change. (Capacitor -> open, Inductor->short). Compute VC or IL using simplified circuit at t=021 August 2005
Engineer M S Ayubi
20
Final Value • Assume circuit has remained in same state for a long time after switch change. (Capacitor -> open, Inductor->short). Compute VC or IL using simplified circuit at t=∞
21 August 2005
Engineer M S Ayubi
21
Time Constant ∀ τ = RC ∀ τ = L/R Units are seconds
21 August 2005
Engineer M S Ayubi
22
General Approach For 0 < t < ∞
Steady State
Transient
Unknown = Final + (Initial – Final)e-t/τ
21 August 2005
Engineer M S Ayubi
23
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
24
Example 1 • Find i(t) through 2 ohm resistor
21 August 2005
Engineer M S Ayubi
25
Example 1: Initial At t=0-, inductor looks like short, but current is zero. i(t= 0-)=0
21 August 2005
Engineer M S Ayubi
26
Example 1: Initial At t=0+, inductor forces current to stay constant i(t= 0+)= i(t= 0-)=0 Initial = 0 A
21 August 2005
Engineer M S Ayubi
27
Example 1: Final At t=∞, inductor again looks like short i(t=∞)=10V / 2Ω = 5A Final = 5A
21 August 2005
Engineer M S Ayubi
28
Example 1: τ τ =L/R RTH seen by inductor is 2 Ω τ = 5H/2 Ω = 2.5 sec
21 August 2005
Engineer M S Ayubi
29
Example 1 i(t) = 5 + (0-5)e-t/τ 6.00
i(t) (Amps)
5.00 4.00 3.00 2.00 1.00 0.00 0
2
4
6
8
10
Time (sec)
21 August 2005
Engineer M S Ayubi
30
Time Constant 120.0%
1-e^-t/tau
100.0% 80.0%
86.5%
95.0%
98.2%
99.3%
99.8%
99.9%
3
4
5
6
7
63.2%
60.0% 40.0% 20.0% 0.0% 1
2
Multiple of tau
21 August 2005
Engineer M S Ayubi
31
Example 2 Find i(t) in left resistor
21 August 2005
Engineer M S Ayubi
32
Example 2: Initial i(t=0-)=10V/(2Ω || 2Ω ) = 10A i(t=0+)= i(t=0-)=10A
Note that current through resistor is 5A at t=0and 10A at t=0+ 21 August 2005
Engineer M S Ayubi
33
Example 2: Final i(t=∞)= 10V / 2Ω = 5A
21 August 2005
Engineer M S Ayubi
34
Example 2: Tau RTH seen by inductor during 0 < t < ∞ is 2 Ω τ = L/R = 2.5 sec
21 August 2005
Engineer M S Ayubi
35
Example 2 i(t) = 5 + (10 – 5)e-t/τ τ = 2.5 sec 11
i(t) (amps)
9
7
5
3 -1
1
3
5
7
9
11
Time (sec)
21 August 2005
Engineer M S Ayubi
36
Example 3 • Find current in left resistor
21 August 2005
Engineer M S Ayubi
37
Game Plan • Part A: Solve problem for 0 < t < 10s – Initial condition? – Final condition?
• Part B: Solve problem for 10 < t < ∞ – Initial condition? – Final condition?
21 August 2005
Engineer M S Ayubi
38
A: Initial Condition • Capacitor looks like open before t=0 • VC(0-)=5A ( 2Ω || 2Ω ) = 5V
21 August 2005
Engineer M S Ayubi
39
A: Initial Condition • VC(0+)= VC(0-)=5V • iR(0+)=5V/2Ω =2.5A
21 August 2005
Engineer M S Ayubi
40
A: Final Condition iR(∞)=5A VC(∞)=10V
21 August 2005
Engineer M S Ayubi
41
A: Tau RTH seen by capacitor = 2 Ω τ = RC = (2Ω ) (3F) = 6 sec
21 August 2005
Engineer M S Ayubi
42
A: Answer For 0 < t < 10 sec iR(t)=5 + (2.5 – 5)e-t/τ τ = 6 sec VC(t)=10 + (5-10) e-t/τ
21 August 2005
Engineer M S Ayubi
43
B: Initial Condition • Initial condition found from previous answer (not steady state) • VC(10-)=10 + (5-10) e-10/τ =9.06V • VC(10+)= VC(10-)=9.06V • iR(10+)=9.1V/2Ω = 4.53V
21 August 2005
Engineer M S Ayubi
44
B: Final Condition iR(∞)=2.5A
21 August 2005
Engineer M S Ayubi
45
B: Tau RTH seen by capacitor = 2Ω || 2Ω = 1Ω τ = RC = (1Ω ) (3F) = 3 sec
21 August 2005
Engineer M S Ayubi
46
B: Answer for 10 < t < ∞ iR(t)=2.5 + (4.53 – 2.5)e-(t-10)/τ τ = 3 sec
21 August 2005
Engineer M S Ayubi
47
Example 3 Graph 5 4.5
i(t)
4 3.5 3 2.5 2 -5
0
5
10
15
20
Time (sec)
21 August 2005
Engineer M S Ayubi
48
Example 4 • Find the voltage across the capacitor Vc(t) for a square wave input (magnitude=10V, period=2ms)
21 August 2005
Engineer M S Ayubi
49
Example 4 • The initial capacitor voltage at t=0 is 0V, heading toward final voltage of 10V (but it will not reach final value before input waveform switches). • The capacitor voltage at t=1ms (which will be between 5 and 10 V) will be the new initial voltage, heading toward a final voltage of 0V (but it will not reach final value before input switches again). 21 August 2005
Engineer M S Ayubi
50
Example 4: Tau τ = RC = (1 KΩ )(1 uF) = 1ms So voltage will get a little over half way each time before the square wave input switches.
21 August 2005
Engineer M S Ayubi
51
Example 4 Pspice Run
21 August 2005
Engineer M S Ayubi
52
Outline • First Order Response – Theory • RL Circuits • RC Circuits
– General Form – Examples
• Second-Order Response – Series RLC – Parallel RLC – General Form 21 August 2005
Engineer M S Ayubi
53
Second Order (RLC)
Series
21 August 2005
Parallel
Engineer M S Ayubi
54
Series RLC
21 August 2005
Engineer M S Ayubi
55
Parallel RLC
21 August 2005
Engineer M S Ayubi
56
General Form 2
d x(t ) dx(t ) a +b + cx(t ) = f (t ) 2 dt dt a b
Series 1 Rth/L
Parallel 1 1/(RthC)
c
1/(LC)
1/(LC)
21 August 2005
Engineer M S Ayubi
57
General Form Steady State
Transient
unknown = final value + natural response
21 August 2005
Engineer M S Ayubi
58
Final Value • Steady state response is considered constant, so use short-cut (capacitor -> open, inductor -> short) to find final value
21 August 2005
Engineer M S Ayubi
59
Natural Response • Guess that transient response is of the form: xtrans(t) = Aept • If constants A and p can be found, then this is the solution
21 August 2005
Engineer M S Ayubi
60
Natural Response 2
d x(t ) dx(t ) a +b + cx(t ) = f (t ) 2 dt dt
ap Ae + bpAe + cAe = 0 2
( ap
pt
pt
2
+ bp + c Ae = 0
)
pt
pt
ap + bp + c = 0 2
21 August 2005
Engineer M S Ayubi
61
Natural Response − b ± b − 4ac p1 , p2 = 2a 2
a, b, and c are circuit parameters. Roots p1, p2 are computed from these parameters. Three possibilities, depending on b2 compared to 4ac 21 August 2005
Engineer M S Ayubi
62
Real Distinct Roots b2 > 4ac Over Damped
xtrans (t ) = A1e
− p1t
+ A2 e
− p2t
Know p1 and p2, must find A1 and A2 from specific solution (using forcing function) 21 August 2005
Engineer M S Ayubi
63
Real Equal Roots b2 = 4ac Critically Damped
xtrans (t ) = A1e
− pt
+ A2te
− pt
Know p= p1=p2, must find A1 and A2 from specific solution (using forcing function)
21 August 2005
Engineer M S Ayubi
64
Complex Roots b2 < 4ac Under Damped
p1, 2 = −α ± jβ xtrans (t ) = Ce
−αt
sin( βt + φ )
Know α and β , must find C and φ from specific solution (using forcing function) 21 August 2005
Engineer M S Ayubi
65
Second Order Response 4 3.5 3
Real Distinct (Over Damped)
2.5 2
Real Equal (Critically Damped)
1.5 1
Complex (Under Damped)
0.5 0 0
10
21 August 2005
20
30
Engineer M S Ayubi
66
Phasor vs. Transient Analysis • Phasor Analysis – source is sine or cosine – no switches
• Transient Analysis – switch and source is constant – square wave can also be treated as a transient
21 August 2005
Engineer M S Ayubi
67
Transient Response • Many other physical systems react with the same type of response (at least approximately) – a first order exponential response starting at an initial condition, moving towards a final condition, with a time constant τ (e.g., temperature probe responding to instantaneous change in temperature)
21 August 2005
Engineer M S Ayubi
68
