450fall09hw2_Sol

# 450fall09hw2_Sol - ECE 450 Fall 2009 Homework 2 Solutions...

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ECE 450 Fall 2009 Homework 2 Solutions Due: Tuesday, Sep 8 th , 2009 1) [] 3 ˆˆˆˆ (,) ( ) ( ) ( ) ( )co s ( ) C t q hx hx hy hy t m ρδ δ ω =− + + + rrrrr , essentially you have four point charges in space that exist in the x - y plane only at points and . ˆ hx ˆ hy a) Since we assume t ρ r is a time harmonic source, i.e., it oscillates at single frequency we can develop the phasor representation as follows: {} { () ' ()co s ( ) '( ) Re Re '( ) Re ( ) jt jt j tt e ee e ωα } ρω + α = + = = = rr r r r ± (1) Thus, r ± is the phasor or complex quantity. The phase of the sinusoid, α , has been absorbed into the phasor and must be accounted for. In t r we see that α =0, therefore 3 ( ) C q h xh yh y m = −−++−−+ r r r ± *To find, t t r , one can easily take the time derivative of t r and find that, 3 ( , ) ( )( ) s i n ( ) tC t q hx hx hy hy t tm δδδδ == + + + r r r r . 1

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In phasor form, one can replace t j ω due to the time harmonic dependence (the time to frequency domain translation for harmonic sources) and simply multiply () ρ r ± by j and bring the expression back into the time domain via the expression in (1). That is, {} [] 3 (,) Re ( ) ˆˆˆˆ s i n ( ) jt t je t C qh xh yh y t m ωρ ωδ δ = = −−++−−+ r r rrrr ± b) Use what you’ve determined above, either in the time domain or phasor domain, respectively that is, ·(,) o r ·() t tj t ∇= = r Jr r ± ± . Therefore, 3 · ( , ) ( )( ) s i n ( ) C t q hx hx hy hy t m = r r r r c) Now we know that each of these four ‘point charges” generate an electric field, associated with the electric field is the scalar potential V . So the problem is as simple as calculating the scalar potential from each of these point charges, and summing the results via superposition. The only difference now is that the charges are time-harmonic in nature hence the notion of finding the ‘retarded’ scalar potential. Start in the phasor domain directly. ' 0 ' 0 4' ( ( ) ' ) 4 jk jk e d x h x h y h y e d π δδδδ −− = = rr r Vr r r ± ± ε ε (2) 2
What we do know, via () ρ r ± , the delta functions tells us exactly where the source locations are located. We have delta functions inside an integral over all space—the only thing to do is to apply the sifting, sampling property to each charge. Recall in rectangular coordinates, ( ) ( ) ˆˆ ˆ ˆ '' ' ' x xy yz z x z −= ++ + + rr Therefore, for example, the first point charge located at ˆ ( hx ) δ r tells us that the charge only exists at , then ˆ ' hx = r ˆ ˆ ' ˆ x z h x = r h x , after applying the sifting property to (2). The rest follows in the same manner. Therefore, ˆˆˆˆ 0 , ˆ ( ˆ 4 ) jk hx jk hx jk hy jk hy qe e e e V hx hx hy hy π −−

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450fall09hw2_Sol - ECE 450 Fall 2009 Homework 2 Solutions...

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