Download presentation
Presentation is loading. Please wait.
1
16.360 Lecture 9 Last lecture Power flow on transmission line
2
16.360 Lecture 9 Last lecture: Power flow on a lossless transmission line Instantaneous power Time-average power i(z) = V(z) = V 0 ( ) + + e jzjz - e -j z (e(e + V0V0 Z0Z0 e jzjz ) At load z = 0, the incident and reflected voltages and currents: V = V 0 + i = i + V0V0 Z0Z0 i V = V 0 - r i = - V0V0 Z0Z0 r
3
16.360 Lecture 9 Instantaneous power + i P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] i i = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] ++ + = (|V 0 |²/Z 0 ) cos²( t + ) + + - r P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] r r = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] +- + = - | |²(|V 0 |²/Z 0 ) cos²( t + + r ) + +
![Lecture 9 Instantaneous power + i P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] i i = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] ++ + = (|V 0 |²/Z 0 ) cos²( t + ) r P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] r r = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] +- + = - | |²(|V 0 |²/Z 0 ) cos²( t + + r ) + +](http://images.slideplayer.com./17/5328635/slides/slide_3.jpg "Lecture 9 Instantaneous power + i P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] i i = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] ++ + = (|V 0 |²/Z 0 ) cos²( t + ) r P(t) = v(t) i(t) = Re[V exp(j t)] Re[ i exp(j t)] r r = Re[|V 0 |exp(j )exp(j t)] Re[|V 0 |/Z 0 exp(j )exp(j t)] +- + = - | |²(|V 0 |²/Z 0 ) cos²( t + + r ) + +")
4
16.360 Lecture 9 Time-average (|V 0 |²/Z 0 ) cos²( t + )dt + + Time-domain approach: P av = i T 1 0 T P (t)dt i = 22 0 T = (|V 0 |²/2Z 0 ) + P av r = -| |² (|V 0 |²/2Z 0 ) + = P av i P av + P av r = (1-| |²) (|V 0 |²/2Z 0 ) + Net average power:
5
16.360 Lecture 9 Time-average Phasor-domain approach P av r = -| |² (|V 0 |²/2Z 0 ) + = (1-| |²) (|V 0 |²/2Z 0 ) + = (½)Re[V i*] P av P av = (1/2) Re[V 0 V 0 * /Z 0 ] i ++ = (|V 0 |²/2Z 0 ) P av
![Lecture 9 Time-average Phasor-domain approach P av r = -| |² (|V 0 |²/2Z 0 ) + = (1-| |²) (|V 0 |²/2Z 0 ) + = (½)Re[V i*] P av P av = (1/2) Re[V 0 V 0 * /Z 0 ] i ++ = (|V 0 |²/2Z 0 ) P av](http://images.slideplayer.com./17/5328635/slides/slide_5.jpg "Lecture 9 Time-average Phasor-domain approach P av r = -| |² (|V 0 |²/2Z 0 ) + = (1-| |²) (|V 0 |²/2Z 0 ) + = (½)Re[V i*] P av P av = (1/2) Re[V 0 V 0 * /Z 0 ] i ++ = (|V 0 |²/2Z 0 ) P av")
6
16.360 Lecture 9 = (½)Re[V i*] P av
![Lecture 9 = (½)Re[V i*] P av](http://images.slideplayer.com./17/5328635/slides/slide_6.jpg "Lecture 9 = (½)Re[V i*] P av")
7
16.360 Lecture 9 Today Smith chart parameter equations input impedance
8
16.360 Lecture 9 Parameter equations. B Open Circuit load Short Circuit load Unit circuit
9
16.360 Lecture 9 Normalized impedance Parameter equations
10
16.360 Lecture 9 Parameter equations
11
16.360 Lecture 9 An example Smith Chart Input impedance Smith Chart Wavelength toward generator (WTG)
12
16.360 Lecture 9 An example Smith Chart find Zin (-0.1 ) Constant | | circle, SWR Circle
13
16.360 Lecture 9 Next lecture: SWR Voltage maxima and minima Impedance to Admittance transformations
14
16.360 Lecture 9 Recall: Smith Chart If when 2 z + r = 2n . | V 0 | [ 1 - | |], + |V(z)| min = when 2 z + r = (2n+1) . | V 0 | [ 1+ | |], + |V(z)| max = SWR, voltage maximum and minimum
Similar presentations
