What is the Limit of Z_n as n Approaches Infinity in Cascaded Resistors?

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The limit of the input impedance of cascaded resistors, defined by the recurrence relation Z_{n+1}=R_1+\frac{R_2 Z_n}{R_2+Z_n}, converges as n approaches infinity. The initial condition is Z_1=R_1+R_2. To find the limit, one can set Z_{n+1} equal to Z_n, simplifying the problem to solving for Z_n in terms of R_1 and R_2. This approach effectively eliminates the need for complex summation and product terms.

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elgen
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Dear all,

The problem of finding the input impedance of a series of cascaded resisters can be stated as

[tex]Z_{n+1}=R_1+\frac{R_2 Z_n}{R_2+Z_n}[/tex] where [tex]Z_1=R_1+R_2[/tex]. What is [tex]\lim_{n\to\infty}Z_n[/tex]?

My attempt is to re-write the recurrance relation as
[tex](R_2+Z_n)Z_{n+1}-(R_1+R_2)Z_n-R_1R_2=0[/tex]

which is
[tex]R_2 Z_{n+1} + Z_{n}Z_{n+1} - (R_1+R_2)Z_n - R_1 R_2 = 0[/tex]

[tex]R_2 Z_{n} + Z_{n-1}Z_{n} - (R_1+R_2)Z_{n-1} - R_1 R_2 = 0[/tex]
...
[tex]R_2 Z_2 + Z_1 Z_2 - (R_1+R_2)Z_1 - R_1 R_2 = 0[/tex]

Summing them up gives
[tex]R_2(Z_{n+1}-Z_1)+Z_{n+1}Z_n + ... + Z_2 Z_1 - R_1(Z_n +...+Z_1) - nR_1R_2=0[/tex].

I am not sure on how to get rid of the product terms and summation terms to get an expression of only [tex]Z_{n+1}, R_1, R_2[/tex] and [tex]n[/tex]. Any suggestion on possible attack?

Thank you.

elgen
 
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hi elgen! :smile:

the question only asks for the limit

you can find the limit (if it exists) by putting Zn+1 = Zn :wink:
 
That sounds cool. Thx for the pointer!
 

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