How to detect telephone status at the other end ?

Hi,

Suppose a call is initated from the other end to me.When I pick my phone I can detect it by the drop in voltage across two telephone lines connected to my phone.I want to know how I can detect the status of the phone at the other end. Currently I'am doing it by detecting the voice signal from the other end when I am not talking but this process needs very clean connection and time. Is there another way I can do this ?

Pubudu

Reply to
Pubudu
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The best way is to change the technology of your incoming phone line.

If your incoming line is POTS (plain old telephone service) that adheres to the North American standard then calls will start with a ring, and when the other end hangs up your line voltage will go to 0V for a significant fraction of a second. Unfortunately, if you have call waiting the second incoming call can cause this switch to 0V also. This is not a bad method -- we used to have an answering machine that used it, and it was flawless until we got call waiting.

If that voltage to 0V doesn't work then this becomes a sort of casual, maybe-it-works-maybe-it-doesn't way of doing things. That's OK for a POTS line because it's designed to have a plain old telephone on the end of the wire with a human on each end, so the protocol included you occasionally saying "Bart! Are you still there!" and whoever's on the other side (hopefully Bart) saying "Yes, I am."

All other styles of phone lines -- T1, ISDN, DID, E & M, etc., are designed to work with machines, and have explicit signaling to let you know what the progress of the call is.

--
Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

I don't think so. Automated telemarketing machines (which are illegal) detect the sound of the word "hello" (which means that if I say "This is ... ..., may I help you?" they get confused); modems detect the sound of a modem; and so forth.

Reply to
mc

Hi Tim,

I dont have a line with POTS facility. I do have a basic telephone line thus I dont see any voltage variation when the other side hooked the phone. For my voice detection, I am not detecting words. I am just detecting a signal. I have a comparator that keep it output at a higher voltage when its input is above the threshold I set. This fails sometimes when the other end have very low volume (circuit stops the conversation) or noisy line (never stops the line).

Pubudu

Reply to
Pubudu

"mc"

** Are they really illegal in the USA ?

Not so here in Aussie.

....... Phil

Reply to
Phil Allison

Pubudu a écrit :

Hello. and if you told us what you have got as telephone installation? By the way POTS is just what it tells and not a "facility". If you have a oldfashioned telephone and not some of the newfangled system, you have POTS. In this case, the connection is made by a current loop (several 10s of mA). This current flows if there is someone on the other side. Conclusion: detect currentflow and not voltage. So, without some more information, I can't say more. Good luck Peter

Reply to
pom

"pom"

** ROTFLMAO !!!!!!..

What does this demented Frog clown think the " other side " is ??

The mythical *After Life * - accessed Ouija boards & Mediums etc ??

Across the famous Ditch, aka the English Channel ??

Dem * current loops " sure must cover some territory .....

....... Phil

Reply to
Phil Allison

Hi Peter, I am in US and my connection is from Verizon. I also have a DSL connection through the same phone line. I can try measuring the current flow because I did not try that method sofar.

em, you have POTS.

What do u mean by this ? I do have a digital phone with answering but I can not tell anything about the line. I think you are talking about the telephone line.

I will check the current flow through the line today.

++++++++++++++++++++++++++++++++++++++++++++++ "Other side " mean the call caller. (call initiater) I am the receiver.

Pubudu

Reply to
Pubudu

"POTS" = basic telephone line.

Not every one of them has the 0V flash on disconnect -- basically you get what you get.

You can get fancy trying to detect cadences or even doing signal processing to detect the spectral features of speech, but no matter what you're doing it can be a pain.

--
Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Plain Old Telephone Service.

Cheers! Rich

Reply to
Rich Grise

If it's through DSL then all bets are off. You need to call your Verizon salesdroids, and try to get through to a tech dept. and ask them these questions.

Careful, though, they might wonder what you're up to!

Good Luck! Rich

Reply to
Rich Grise

As has been pointed out, there are a number of states which may indicate a distant party hang-up, depending on your switch type, etc. Momentary voltage drops, polarity reversals, etc. are all possibilities. You could try calling someone or getting them to call, observing any change in the line's electrical properties. The main problem with electrical-based detection methods stems from the fact that the same states might occur under other conditions. For example, in the UK, this is signalled by a steady 400Hz NU tone, which is also used to signal other things (such as Call Waiting) which means the detection circuit must make sure that the NU tone is of an appropriate duration for a hang-up signal.

HTH,

Neil

Reply to
neil90s

From my experience there is no DC or other line signal which passes through to the called (B) party, to indicate whether the A party is looped or not, other than speech in the first case, or congestion tone in the latter. The latter occurs if the A party hangs up while the B party remains looped. The A party hanging up will initiate a disconnect condition to the originating exchange and the B party receives congestion tone from the terminating exchange. Conversely, if the A (calling) party remains looped and the B party hangs up, a time-out period (IIRC it used to be around 30 seconds) will commence to signal to the A party with congestion tone that the B party has hung up. At least that's how it worked with L M Ericsson x-bar but it is probably a bit different with modern digital exchanges.

Reply to
Ross Herbert

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