Complex Numbers, right track?

Discussion in 'Math' started by Biggsy100, Jun 8, 2015.

  1. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    I have a question relating to complex numbers that I need my workings clarified:

    A transmission line under tes has the following primary constants:

    L= 125 X 10^-9 Henries per metre
    R= 3 X 10^-3 Ohms per metre
    G= 1 X 10^-7 per metre
    C= 25 X 10^-12 Farads per metre.

    Calculate the magnitude of the characteristic impedance with a 5kHz line signal frequency. Comment on the result and mention any reasonable assumption made in your calculation.

    So I know that C/Impedance of a transmission line is given by the equation: Z^o = R+jωL/G+jωC

    (3x10^-3) + (5x125x10^-9)/ (1x10^-7)+(5 x 25 x10^-12) = 30586.76

    Am I right to think the j is = the 5kHz? Also, the intergration of ω? Is that just a continuation?
  2. WBahn


    Mar 31, 2012
    a + b/c + d = a + (b/c) + d and NOT (a+b)/(c+d)

    You also need to track your units.
  3. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    Wow? sO, How do I know which is A and B if they are all given different letters?
  4. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    So is it?

    3x10^-3 + (5/125x10^-9)+125x10^-9 = 2x10^11 ?
  5. WBahn


    Mar 31, 2012
    I don't follow. What A and B are you referring to?

    The equation you say that you are starting from is:

    Z^o = R+jωL/G+jωC

    Because multiplication and division take precedence over addition and subtraction, what you are actually saying is that this is

    Z^o = R + (jωL/G) + jωC

    which is incorrect. What you meant so say was

    Z^o = (R+jωL)/(G+jωC)

    since you need to do the addition first, so you need to use parentheses in order to force that to happen.

    The frequency, f, of 5kHz is the frequency in cycles per second. What you need is ω which is the frequency in radians per second. There are 2pi radians in one cycle, so

    ω = 2*pi*f = 10*pi rad/sec

    The j is a constant, specifically the number that, when squared, equals -1.

    j = √(-1)

    This is known as the "imaginary constant". The impedance (a generalization of resistance) of an inductor or capacitor is purely imaginary, which simply means that it is a "normal" or "real" number multiplied by the imaginary constant. A general impedance, such as R+jωL, has both a real part and an imaginary part -- this is known as a complex number.

    When working with complex numbers, you treat the j like a constant parameter (which is exactly what it is) just like you normally would in algebra. The only difference is that any time you end up with j² you can replace that with -1.

    In general, the characteristic impedance of a transmission line is a complex number.

    Tracking your units is very important. Had you done that, you would have seen that your result would have ended up with inconsistent units in three places. First, the way you evaluated jωL would not have yielded a quantity with units of resistance (known as impedance in the complex case), which is required in order to add that quantity to R. Second, the way you evaluated jωC would not have yielded a quantity with units of conductance (known as admittance in the complex case), which is required in order to add that quantity to G. Finally, even if these errors hadn't been made, you would have ended up with (impedance)/(admittance) which has units of (impedance-squared) when you want just units of impedance. This would have highlighted that there was a problem with your starting equation. The equation you wanted was:

    <br />
Z_0 \; = \; \sqrt{\frac{R + j \omega L}{G + j \omega C}}<br />
<br />
    MrAl likes this.
  6. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    Ok, that's a lot clearer for me to understand, however now inputting numbers into my calculator I get a Syntax error:

    √(3x10^-3 + 2*pi*5 x 125x10^-9)^-1 / (1x10-7+ 2*pi*5 x 125x10^-9)^-1
  7. MrAl

    Well-Known Member

    Jun 17, 2014

    Hello there,

    When you work with complex numbers the 'j' refers to the imaginary part of the number, so a complex number with real part 3 and imaginary part 8 can be written as:

    The magnitude is then:

    and the phase is:

    You dont seem to have much experience with complex numbers so i would suggest that you look into this subject as a prerequisite before you work with circuits that require complex math to solve.

    What this means is that you should get both a magnitude and phase angle for the impedance. If you dont know what this means then you'll need to look into this subject area first or none of the results you get will make any sense.

    Also, a rough rough estimate of the impedance when the frequency is high is:

    but that's only when the reactances are high compared to the resistance and conductance, which usually occurs at high frequency.
    Last edited: Jun 9, 2015
  8. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    So it is separate sums and then these values are entred into the final equation.

    FYI- I am undertaking a general engineering course, covering both Electrical and Mechanical. I am Mechanically biased so Electronics is not my forte.
  9. WBahn


    Mar 31, 2012
    Why are you raising terms to the -1 power?

    I don't know what the order of operations are on your calculator for square root.

    But even if you aren't getting a syntax error, you aren't doing it right. The quantity j is equal to the square root of minus one. You don't have that anywhere in your expression and, even if you did, almost no calculators handle complex numbers (though that might not be the case with modern scientific calculators).

    You really need to spend a bit of time learning how to work with complex numbers. The basics are pretty straightforward, but you need to go a bit beyond the basics to do the square root since that is most easily done using exponential notation.

    Here's a site that might be helpful.

    I put this together about fifteen years ago and a lot of people have found it useful over the years.
  10. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    I have looked into complex numbers extensively, still very stuck. It's only easy if you know the answer. I don't know the answer so it's not easy
    Last edited: Jun 9, 2015
  11. WBahn


    Mar 31, 2012
    That's why you need to take things from square one. Pick a tutorial or other material and work through it step by step. At each point you get stuck, after struggling with it for a while, ask for help.
  12. MrAl

    Well-Known Member

    Jun 17, 2014

    That's ok, both share much of the same kind of math, so learning the math is basic to both disciplines.

    Complex math is not that hard, but you really should know algebra fluently first. I have a feeling you havent done much with algebra yet either. For example, you cant write:
    when you really mean:

    You really need to get that straight.

    Once you get that down, complex math is not that much different, there are just a few more rules to know. These few rules will get you going in no time.

    We've all seen multiplication of two numbers like 5 and 7, and know that 5*7 equals 35. We learned that long time ago. Now we have complex math so we need to update those ideas for addition, subtraction, multiplication and division.

    Complex numbers have at least two parts, the real and imaginary parts. When we add two complex numbers we add the two parts of each number. So for example when we add:
    1+2j to 3+4j
    we add the 1 to the 3 and get 4, and add the 2 to the 4 and get 6, so we end up with:
    See how simple that is?

    Now if we multiply those two instead, we have to follow the rules of algebra as if 'j' was just a variable that was as of yet undetermined. So lets look first at multiplying these two:

    How would you multiply these two as:

    The answer is simple. You multiply just the same as any other algebraic expression, and that means multiply 1 times 3 and 1 times 4x and 2x times 3 and 2x times 4x, and that gives us four terms:

    You should have no problem with this so far otherwise you need to review algebra first.

    Now we can simplify that into:

    And that's the end result when we do it as a simple algebraic expression. But in our original multiplication x was really the complex operator 'j', so we replace each x with j and get:

    and we are almost done. We need a couple more identities before we can complete this. These are the powers of 'j'. A Short table of powers of 'j':

    and after that the pattern repeats so we have:

    and the same pattern repeats for powers above 8 also.
    Note that sometimes we end up with a real number and sometimes not.

    Now back to our previous result:

    and we see one power of 'j', so we replace that with it's equivalent of -1 from the table and we get:


    and now we can combine the 3 and -8 and get:

    as the final result of the multiplication of the two complex numbers far above.

    So now you should know how to add and multiply, and for division Wbahn showed you how to do that which i'll repeat here briefly. To divide:

    multiply top and bottom by the conjugate of the denominator 3+4j which comes out to 3-4j and then simplify the result.

    Subtraction is the same as addition except you subtract the two individual parts. Sometimes you'll get a completely real number and sometimes a completely imaginary number:
    (2+3j)-(1+3j)=1+0j=1 (real)
    (2+3j)-(2-3j)=0+6j=6j (imaginary)

    So using complex numbers really isnt that much harder than using regular numbers, there are just a few more steps sometimes and a few more rules to learn. If you use software to do any calculations that software has to be capable of handling complex numbers too or else you have to calculate the real and imaginary parts individually.

    There are also rules for functions like cos(X), sin(X), ln(X), X^Y, etc., which you can learn later on.
    Last edited: Jun 10, 2015