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Conquer Radio Frequency

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2.14 The dreaded maths 119 ( ) ( ) ( ) ( ) ( ) ( ) In this case, where we assume that no losses are incurred along the line, the solution of (2.14-8) for the voltage will be of the form 25 shown in (2.14-9) ( ) [ ] The and components of this equation represent the incident and reflected voltages respectively. This is the equation which is used to model the incident and reflected waveforms which we saw in section 2.8. For the current we get a similar solution as shown by eq. (2.14-10) ( ) [ ] [ ] Again the and component represent the incident and reflected currents respectively. Remember that ⁄ √ , where represents the effective wavelength inside the line (section 2.11, eq. (2.11-3)) As mentioned in section 2.3, eq. (2.3-4), in the lossless case may also be expressed as √ is related to the amplitudes of incident and reflected voltages and currents by equation (2.14-11). The minus for the backward wave is significant and is indicative of the fact that the current is moving in the negative direction, away from the load. Now, to simplify our analysis, we may assume that we are in sinusoidal steady-state 26 at angular frequency and hence omit the term. This makes (2.14-8) a bit simpler as shown by (2.14-12). ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 25 We are using the complex exponential representation for our solution as we did in (2.14-1) however the time domain solution will be similar to (2.14-5) i.e. { ( )} ( ) ( ) 26 Sinusoidal steady-state refers to the steady-state that gets established in a circuit when all the independent sources are sinusoids of the same angular frequency (2.14-8) (2.14-9) (2.14-10) (2.14-11) (2.14-12) Conquer Radio Frequency 119 www.cadence.com/go/awr

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