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Help me match impedance between the coaxial and speaker

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susannit

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hi,
I need to match a coaxial cable of 50ohm impedance to a speaker of 8ohm. kindly tell for matching the impedance between the coaxial and speaker.
 

impedance matching formula

An ordinary audio/music speaker? A conventional audio power amp on the other end? You can connect the coax directly to the speaker and amp. Cable impedance matching is not required or desirable in this situation.

Why not use ordinary speaker wire? It's cheaper and has lower resistance and lower capacitance than most coax.

By the way, a speaker may say "8 ohm", but it usually has a wildly complex impedance. For example, here is an impedance plot of a "4 ohm" hi-fi speaker:
#528088
 

    susannit

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speaker complex impedance

When you have a transmission line and a load at the far end, you usually only care to match their impedances when the electrical line length is large. And this is because for large electrical line lengths, you will get return reflections that can vectorialy cancel your signal, and cause odd effects. Electrically short line lengths do not allow the reflected waves to have enough phase shift to cause any cancellation. By "electrical line length", I mean the physical length of the transmission line in relationship to a wavelength.

Since you are using a speaker, I would assume that you are sending audio signals to the speaker. The highest signal will be 20 KHz. Lets say the speaker is 100 meters away from the amplifier. The wavelength at 20 KHz is:

λ = C/f (where C is the speed of light, and f is the frequency in Hz)

so λ = (3 * 10^8 m/s) / (2 * 10^4 Hz) = 1.5 * 10^4 m =15000 meters

If your speaker/amplifier spacing was 100 meters, you have an electrical line length of 1/150 of a wavelength, which is very small.

So unless your physical spacing is some significant fraction of 15 killometers, you do not have to worry about impedance matching the load to the coaxial cable. Your system is acting like they are "lumped elements", not "distributed elements".

(BTW, units of Hz are the same as units of 1/seconds, which is why they cancel out)
 

    susannit

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complex impedance speaker wire

echo47 said:
An ordinary audio/music speaker? A conventional audio power amp on the other end? You can connect the coax directly to the speaker and amp. Cable impedance matching is not required or desirable in this situation.

Why not use ordinary speaker wire? It's cheaper and has lower resistance and lower capacitance than most coax.

By the way, a speaker may say "8 ohm", but it usually has a wildly complex impedance. For example, here is an impedance plot of a "4 ohm" hi-fi speaker:
#528088

And RG58 (one kind of 50 Ohm cable) itself have around 6390 Ohm and -45 degree characteric complex impedance on 20 Hz, 1013 Ohm and -44.92 degree on 800 Hz etc.


Is not easy task to make power matched coupling with near zero Ohm impedance in amplifier, up to 6.3 kOhm complex characteric impedance in cable (20 Hz) and complex 4-60 Ohm complex impedance on speaker... and you not want this if you want well controlled speaker diaphragma moving... good matching is same as lose control over load side from amplifier side ie. not make reflection.

50, 75, 100 Ohm characteric impedance on coax, pair and twisted cable is only valid for frequency >= 1 MHz, same cable have characteric complex impedance around 4-6 kOhm on 20 Hz, 600-900 Ohm on 1000 Hz and 200-300 Ohm in 10 kHz and all have around -45 degree phase angle reactive component ('-' sign on angle indicate capacitive load)

fortunly - influenses of cable impedance itself in speaker uses is very small to not possibly to measure for distance very much shorter than electrical wavelength on cable for used frequency (around 384 km for 20 Hz or 7684 km/s - 0.025c - Yes, low frequency electromagnetic wave travel very slowly on cable depend of dominating RCRCRCRC-chain and result very small voltage difference and long charging time each infinite small RC-segment in cable. Higher freqency - faster travel (but more loss) and frequency as 10 kHz have around 0.45 of speed of light)

Biggest effect in short distance case is a serie resistance in cable compare to low impedance load (speaker) to delevery much power, and also absorb power from speakers moving mass to controll diaphragma moving, and cable capacitance if source have high impedance (microphone).
 

    susannit

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long transmission linesmatching

xxargs said:
echo47 said:
An And RG58 (one kind of 50 Ohm cable) itself have around 6390 Ohm and -45 degree characteric complex impedance on 20 Hz, 1013 Ohm and -44.92 degree on 800 Hz etc.

).

You lost me on that one. Coaxial cable characteristic impedance is ~not frequency dependent. A 50 ohm cable at 100 MHz is, surprise, still 50 ohm cable at 1 KHz. Impedance has to do with √(L/C), where L is inductance per length, and C is capacitance per length. Does not change with frequency.
 

    susannit

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power line impedance matching

biff44 said:
[Edit missing removed quote]


You lost me on that one. Coaxial cable characteristic impedance is ~not frequency dependent. A 50 ohm cable at 100 MHz is, surprise, still 50 ohm cable at 1 KHz. Impedance has to do with √(L/C), where L is inductance per length, and C is capacitance per length. Does not change with frequency.

Ah, You also read and use textbooks simplified formula!!!

textbook have written mostly two simplified case:

for audio frequency 0 - ~20 kHz) ie. g << ωr and r >> ωl is simplified to

γ = α + jβ = √(jωrc)


α = β = √((ωrc)/2) or γ = √(ωrc) |_ 45 degree

Z = √(r/(jωc) or |Z| = √(r/(ωc)) and |_- 45 degree

α is attenuate in Neper per unit length
β i phase constant i radianer per sec and unit length

(this works up ~ 10 KHz on ordinary twisted pair but angle is not longer inside 44-45 degree range for frequency above 10 KHz if calculate with full telegraph formula down here)

For high frequency (>= 1 MHz) ie. g << ωc and r << ωl is simplified formula as:

α = (r/2 * √(c/l)) + (g/2 * √(l/c))

β = ω√(lc)

Z = √(l/c) and asumme is a real impedance


---

Above is simplified near value formule make from complete telegraph equation as:

γ = α + jβ = √((r + jwl) * (g + jωc))

Z = √((r + jωl) / (g + jωc))


Most Textbook make lot of paperspace to describe how to simplifie this formula to above simplified near vaule formula so people not need solve (to much) complex equtions...

But near value formula handle around 0-10 KHz and 1 MHz and up but still miss some case intressting range between 10 kHz to 1 MHz (ie ADSL frequency) - most textbook skip this part without comments...

---

If you have calculators can solve complex square root and easly handle +,-,/,* with complex number (ex. HP42S), is a no reason to use simplified formula first part in this text. I'm always using full telegraf equation to done this, is more easy to remember and give more exact result compare to near value formula. (unfortly, excel:s 'IMPOWER' cannot handle power of 0.5 (to make complex square root) or complex power (e^a+ib - using) on complex number, only allow integer... - why not written complete complex support in excel if still try written some complex functions, Microsoft??? )

For example simplified formula can make value, using in other formula, give cable allow signal going faster than speed of light without notice and indicate simplified formula used on outside valid range...

---

Value 'g' is near always missing or unknow - so you can try '0' on low frequency or if unknow plastic as cable insulation and higher frequency, you can try g = ω * c * δ and δ = 0.0003 or look up in table for used insulations material

δ =tan(d) = DF (Dissipations Factor) = 1/Q = esr/Xc

DF can written as 0.035 or 3.5 %

ex: Tan(0.01719 degree) = DF 0.0003 = DF 0.03%

you need also adjust value 'r' depend of frequency on high frequency depend of skin effect on conductor as r = constant * √(f) above > 50 KHz - constant is very depend of used cable geometry and used metal in conductor.
 

audio 50 ohm to 600 ohm impedance match

Please be careful with your quotations. I did not write some of the words labeled "echo47 wrote:"
 

Re: impedance matching.

Sorry, the website added that "echo47 wrote", not me.

Xargs: Hmmmm. I see what you are saying. Because ω is small at audio frequencies, the the approximation that Z=√L/C does not apply, and Z is actually complex, and dominated by the R and G elements, as suggested by the telegrapher's equation. Since a typical line has parameters in the 10 nH/meter and 20 pF/meter range, I can see your point.

I will have to think on this one for a while though. All my experience, with things like terminating cables in 50 ohms at an oscilloscope to reduce triple travel ringing, says that even at low frequencies a 50 ohm cable has 50 ohm impedance. But, such rings may be up in the MHz range, not in the Hz range.

There is a large body of work with power distribution grids and the reflection of waves on very long transmission lines, due to lightning strikes, etc, which are at very low frequencies. I will try to find some corroborating stuff one way or the other.

Very interesting.
 

Re: impedance matching.

Air hanged big power plant lines and old air hanged phone/telegraph line have different situation compare to compact cable - this line have big space between conductor and make higher inductance and lower capacitanse - so this line going to Z= √L/C-state already on audio frequencys as 50/60 Hz and 300- 3400 Hz as phone.

and remember - power plant line is not power matched system to matched load, is optimized to make lowest possible loss in transmission link - but still must handle all type of effect and time delayed reflections coming back from transmissions system and all types of in many case complex terminate impedance of different value over time to hold power line in stable state (simular loudspeaker system)

powerplant line have impedance around 900 - 1200 Ohm and air hanged phone line with 3 mm copper and ~0.4m spacing have 600 Ohm real impedance and bandwidh couple of MHz, and is why all LF-levels always refered to 600 Ohm impedance reference from this time.

Unknow differences between air hanged line with real impedance and compact cable with complex impedance on audio frequency make heavy loss for investors in couple of Atlantic cable installation between USA and Europa in begining of 1900:s - before to discover telegrapher's equation, and understod why this make heavy high frequency loss (we talk 'high frequency' as above few Hz - not kHz) and make insigth to need using pupin solve with extra inductance in serie each cable segment - even i first look seem insane stupid solution...

is a old picture to show this effect - only for 100 km distance :

www.algonet.se/~toek/pupin.GIF
 

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